Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 1)^{2}
= x^{4} + 16x^{3} + 37x^{2} 126x + 72 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 1)^{2} = x^{4} + 16x^{3} + 37x^{2} 126x + 72
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,1372), (10,968), (7,320), (4,400), (1,220), (2,112), (5,2992) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a double root at x = 1. Not just the function but also its first derivative are zero at this point. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 1)^{2} = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 48x^{2} + 74x 126
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 48x^{2} + 74x 126 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.7, 3.2, 1.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 96x^{2} + 37x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 7.14 and 0.86. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 1)^{2}
= x^{4} 8x^{3} 59x^{2} + 138x 72 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 1)^{2} = x^{4} 8x^{3} 59x^{2} + 138x 72
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,1216), (4,800), (1,260), (2,80), (5,1232), (8,2744), (11,1700), (14,6760), (17,29440) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a double root at x = 1. Not just the function but also its first derivative are zero at this point. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 1)^{2} = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 24x^{2} 118x + 138
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 24x^{2} 118x + 138 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.8, 1.0, 8.9]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 48x^{2} 59x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 1.72 and 5.72. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x+13)(x 1)
= x^{4} + 30x^{3} + 275x^{2} + 630x 936 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x+13)(x 1) = x^{4} + 30x^{3} + 275x^{2} + 630x 936
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,240), (11,120), (8,360), (5,336), (2,1320), (1,0), (4,8160) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 13. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x+13)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 90x^{2} + 550x + 630
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 90x^{2} + 550x + 630 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [12.5, 8.4, 1.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 180x^{2} + 275x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 10.73 and 4.27. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x+13)(x 1)
= x^{4} + 6x^{3} 157x^{2} 786x + 936 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x+13)(x 1) = x^{4} + 6x^{3} 157x^{2} 786x + 936
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,3120), (11,2760), (8,1800), (5,816), (2,1848), (1,0), (4,4080), (7,7800), (10,6624), (13,5928), (16,38280) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 13. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x+13)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 18x^{2} 314x 786
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 18x^{2} 314x 786 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.2, 2.3, 8.2]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 36x^{2} 157x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.83 and 3.83. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x+13)(x 4)
= x^{4} + 27x^{3} + 182x^{2} 288x 3744 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x+13)(x 4) = x^{4} + 27x^{3} + 182x^{2} 288x 3744
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,288), (11,150), (8,480), (5,504), (2,2640), (1,3822), (4,0), (7,14820) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 13. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x+13)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 81x^{2} + 364x 288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 81x^{2} + 364x 288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [12.5, 8.4, 0.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 162x^{2} + 182x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 10.65 and 2.85. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x+13)(x 4)
= x^{4} + 3x^{3} 178x^{2} 336x + 3744 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x+13)(x 4) = x^{4} + 3x^{3} 178x^{2} 336x + 3744
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,3744), (11,3450), (8,2400), (5,1224), (2,3696), (1,3234), (4,0), (7,3900), (10,4416), (13,4446), (16,30624) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 13. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x+13)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 9x^{2} 356x 336
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 9x^{2} 356x 336 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.1, 0.9, 8.9]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 18x^{2} 178x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.25 and 4.75. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x+13)(x+5)
= x^{4} + 36x^{3} + 461x^{2} + 2466x + 4680 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x+13)(x+5) = x^{4} + 36x^{3} + 461x^{2} + 2466x + 4680
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,144), (11,60), (8,120), (5,0), (2,1320) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 13. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x+13)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 108x^{2} + 922x + 2466
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 108x^{2} + 922x + 2466 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [12.5, 9.0, 5.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 216x^{2} + 461x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 11.04 and 6.96. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x+13)(x+5)
= x^{4} + 12x^{3} 115x^{2} 1686x 4680 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x+13)(x+5) = x^{4} + 12x^{3} 115x^{2} 1686x 4680
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,1872), (11,1380), (8,600), (5,0), (2,1848), (1,6468), (4,12240), (7,15600), (10,11040), (13,8892), (16,53592) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 13. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x+13)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 36x^{2} 230x 1686
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 36x^{2} 230x 1686 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.6, 5.4, 7.2]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 72x^{2} 115x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.31 and 2.31. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 1)(x 4)
= x^{4} + 13x^{3} 14x^{2} 288x + 288 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 1)(x 4) = x^{4} + 13x^{3} 14x^{2} 288x + 288
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,1666), (10,1232), (7,440), (4,640), (1,550), (2,224), (5,748), (8,7840) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 1. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 1)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 39x^{2} 28x 288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 39x^{2} 28x 288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.7, 2.7, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 78x^{2} 14x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.84 and 0.34. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 1)(x 4)
= x^{4} 11x^{3} 38x^{2} + 336x 288 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 1)(x 4) = x^{4} 11x^{3} 38x^{2} + 336x 288
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,1672), (4,1280), (1,650), (2,160), (5,308), (8,1568), (11,1190), (14,5200), (17,23920) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 1. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 1)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 33x^{2} 76x + 336
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 33x^{2} 76x + 336 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.5, 2.6, 9.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 66x^{2} 38x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.98 and 6.48. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 15)(x 1)
= x^{4} + 2x^{3} 201x^{2} 882x + 1080 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 15)(x 1) = x^{4} + 2x^{3} 201x^{2} 882x + 1080
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,2744), (10,2200), (7,880), (4,1520), (1,1760), (2,1456), (5,7480), (8,13720), (11,15640), (14,6760), (17,21344), (20,79040) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 15. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 15)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 6x^{2} 402x 882
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 6x^{2} 402x 882 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.5, 2.2, 10.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 12x^{2} 201x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.31 and 5.31. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 15)(x 1)
= x^{4} 22x^{3} + 39x^{2} + 1062x 1080 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 15)(x 1) = x^{4} 22x^{3} + 39x^{2} + 1062x 1080
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,3344), (4,3040), (1,2080), (2,1040), (5,3080), (8,2744), (11,680), (14,520), (17,3680), (20,19760) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 15. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 15)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 66x^{2} + 78x + 1062
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 66x^{2} + 78x + 1062 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.2, 6.1, 13.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 132x^{2} + 39x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.63 and 10.37. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 15)(x 4)
= x^{4} 1x^{3} 210x^{2} 288x + 4320 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 15)(x 4) = x^{4} 1x^{3} 210x^{2} 288x + 4320
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,3332), (10,2800), (7,1210), (4,2432), (1,4400), (2,2912), (5,1870), (8,7840), (11,10948), (14,5200), (17,17342), (20,66560) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 15. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 15)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 3x^{2} 420x 288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 3x^{2} 420x 288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.5, 0.6, 11.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 6x^{2} 210x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.67 and 6.17. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 15)(x 4)
= x^{4} 25x^{3} + 102x^{2} + 1008x 4320 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 15)(x 4) = x^{4} 25x^{3} + 102x^{2} + 1008x 4320
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,4598), (4,4864), (1,5200), (2,2080), (5,770), (8,1568), (11,476), (14,400), (17,2990), (20,16640) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 15. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 15)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 75x^{2} + 204x + 1008
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 75x^{2} + 204x + 1008 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [2.4, 7.6, 13.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 150x^{2} + 102x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 1.55 and 10.95. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 15)(x+5)
= x^{4} + 8x^{3} 183x^{2} 2070x 5400 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 15)(x+5) = x^{4} + 8x^{3} 183x^{2} 2070x 5400
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,1568), (10,1000), (7,220), (4,304), (1,3520), (2,10192), (5,18700), (8,25480), (11,25024), (14,9880), (17,29348), (20,104000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 15. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 15)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 24x^{2} 366x 2070
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 24x^{2} 366x 2070 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.9, 5.4, 9.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 48x^{2} 183x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 7.87 and 3.87. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 15)(x+5)
= x^{4} 16x^{3} 87x^{2} + 1170x + 5400 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 15)(x+5) = x^{4} 16x^{3} 87x^{2} + 1170x + 5400
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,836), (4,608), (1,4160), (2,7280), (5,7700), (8,5096), (11,1088), (14,760), (17,5060), (20,26000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 15. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 15)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 48x^{2} 174x + 1170
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 48x^{2} 174x + 1170 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 3.9, 13.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 96x^{2} 87x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 1.52 and 9.52. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 1)(x+5)
= x^{4} + 22x^{3} + 139x^{2} + 198x 360 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 1)(x+5) = x^{4} + 22x^{3} + 139x^{2} + 198x 360
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,784), (10,440), (7,80), (4,80), (1,440), (2,784), (5,7480) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 1. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 1)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 66x^{2} + 278x + 198
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 66x^{2} + 278x + 198 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.1, 5.5, 0.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 132x^{2} + 139x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.16 and 2.84. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 1)(x+5)
= x^{4} 2x^{3} 101x^{2} 258x + 360 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 1)(x+5) = x^{4} 2x^{3} 101x^{2} 258x + 360
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,304), (4,160), (1,520), (2,560), (5,3080), (8,5096), (11,2720), (14,9880), (17,40480) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 1. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 1)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 6x^{2} 202x 258
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 6x^{2} 202x 258 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 1.3, 8.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 12x^{2} 101x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.63 and 4.63. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x+12)(x 1)
= x^{4} + 23x^{3} + 156x^{2} + 252x 432 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x+12)(x 1) = x^{4} + 23x^{3} + 156x^{2} + 252x 432
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,686), (10,352), (7,40), (4,160), (1,550), (2,896), (5,8228) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 12. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x+12)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 69x^{2} + 312x + 252
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 69x^{2} + 312x + 252 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.2, 6.0, 1.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 138x^{2} + 156x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.41 and 3.09. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x 12)(x 1)
= x^{4} 1x^{3} 108x^{2} 324x + 432 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x 12)(x 1) = x^{4} 1x^{3} 108x^{2} 324x + 432
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,152), (4,320), (1,650), (2,640), (5,3388), (8,5488), (11,2890), (14,10400), (17,42320) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 12. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x 12)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 3x^{2} 216x 324
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 3x^{2} 216x 324 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 1.6, 8.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 6x^{2} 108x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 4.0 and 4.5. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x+12)(x 4)
= x^{4} + 20x^{3} + 84x^{2} 288x 1728 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x+12)(x 4) = x^{4} + 20x^{3} + 84x^{2} 288x 1728
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,833), (10,448), (7,55), (4,256), (1,1375), (2,1792), (5,2057), (8,15680) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 12. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x+12)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 60x^{2} + 168x 288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 60x^{2} + 168x 288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.1, 6.0, 1.2]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 120x^{2} + 84x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.32 and 1.68. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x 12)(x 4)
= x^{4} 4x^{3} 108x^{2} + 1728 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x 12)(x 4) = x^{4} 4x^{3} 108x^{2} + 1728
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,209), (4,512), (1,1625), (2,1280), (5,847), (8,3136), (11,2023), (14,8000), (17,34385) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 12. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x 12)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 12x^{2} 216x + 0
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 12x^{2} 216x + 0 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 0.0, 9.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 24x^{2} 108x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.36 and 5.36. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x+12)(x+5)
= x^{4} + 29x^{3} + 300x^{2} + 1332x + 2160 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x+12)(x+5) = x^{4} + 29x^{3} + 300x^{2} + 1332x + 2160
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,392), (10,160), (7,10), (4,32), (1,1100) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 12. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x+12)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 87x^{2} + 600x + 1332
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 87x^{2} + 600x + 1332 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.4, 6.0, 5.3]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 174x^{2} + 300x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.85 and 5.65. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x 12)(x+5)
= x^{4} + 5x^{3} 108x^{2} 972x 2160 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x 12)(x+5) = x^{4} + 5x^{3} 108x^{2} 972x 2160
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,38), (4,64), (1,1300), (2,4480), (5,8470), (8,10192), (11,4624), (14,15200), (17,58190) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 12. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x 12)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 15x^{2} 216x 972
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 15x^{2} 216x 972 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 5.3, 7.6]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 30x^{2} 108x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.67 and 3.17. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 8)(x 1)
= x^{4} + 9x^{3} 82x^{2} 504x + 576 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 8)(x 1) = x^{4} + 9x^{3} 82x^{2} 504x + 576
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,2058), (10,1584), (7,600), (4,960), (1,990), (2,672), (5,2244), (8,0), (11,11730) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 8. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 8)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 27x^{2} 164x 504
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 27x^{2} 164x 504 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.6, 2.4, 5.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 54x^{2} 82x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.58 and 2.08. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 8)(x 1)
= x^{4} 15x^{3} 10x^{2} + 600x 576 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 8)(x 1) = x^{4} 15x^{3} 10x^{2} + 600x 576
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,2280), (4,1920), (1,1170), (2,480), (5,924), (8,0), (11,510), (14,3120), (17,16560) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 8. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 8)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 45x^{2} 20x + 600
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 45x^{2} 20x + 600 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.3, 4.3, 10.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 90x^{2} 10x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.22 and 7.72. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 8)(x 4)
= x^{4} + 6x^{3} 112x^{2} 288x + 2304 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 8)(x 4) = x^{4} + 6x^{3} 112x^{2} 288x + 2304
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,2499), (10,2016), (7,825), (4,1536), (1,2475), (2,1344), (5,561), (8,0), (11,8211) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 8. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 8)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 18x^{2} 224x 288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 18x^{2} 224x 288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [9.5, 1.2, 6.3]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 36x^{2} 112x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.07 and 3.07. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 8)(x 4)
= x^{4} 18x^{3} + 32x^{2} + 672x 2304 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 8)(x 4) = x^{4} 18x^{3} + 32x^{2} + 672x 2304
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,3135), (4,3072), (1,2925), (2,960), (5,231), (8,0), (11,357), (14,2400), (17,13455) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 8. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 8)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 54x^{2} + 64x + 672
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 54x^{2} + 64x + 672 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [2.7, 5.9, 10.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 108x^{2} + 32x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.64 and 8.36. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+12)(x 8)(x+5)
= x^{4} + 15x^{3} 22x^{2} 936x 2880 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x 8)(x+5) = x^{4} + 15x^{3} 22x^{2} 936x 2880
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(13,1176), (10,720), (7,150), (4,192), (1,1980), (2,4704), (5,5610), (8,0), (11,18768) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 8. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+12)(x 8)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 45x^{2} 44x 936
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 45x^{2} 44x 936 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.0, 5.4, 4.3]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 90x^{2} 22x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 7.96 and 0.46. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 12)(x 8)(x+5)
= x^{4} 9x^{3} 94x^{2} + 456x + 2880 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 12)(x 8)(x+5) = x^{4} 9x^{3} 94x^{2} + 456x + 2880
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,570), (4,384), (1,2340), (2,3360), (5,2310), (8,0), (11,816), (14,4560), (17,22770) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 12. There is a single, unique root at x = 8. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 12)(x 8)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 27x^{2} 188x + 456
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 27x^{2} 188x + 456 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 2.1, 10.3]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 54x^{2} 94x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 2.3 and 6.8. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 1)^{2}
= x^{4} + 8x^{3} + 5x^{2} 38x + 24 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 1)^{2} = x^{4} + 8x^{3} + 5x^{2} 38x + 24
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,192), (4,0), (1,60), (2,48), (5,1584) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a double root at x = 1. Not just the function but also its first derivative are zero at this point. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 1)^{2} = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 24x^{2} + 10x 38
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 24x^{2} + 10x 38 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.1, 1.8, 1.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 48x^{2} + 5x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.78 and 0.22. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x 1)^{2}
= x^{4} 27x^{2} + 50x 24 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x 1)^{2} = x^{4} 27x^{2} + 50x 24
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,704), (4,400), (1,100), (2,16), (5,176), (8,2744) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a double root at x = 1. Not just the function but also its first derivative are zero at this point. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x 1)^{2} = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 54x + 50
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 54x + 50 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [4.0, 1.0, 3.1]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 27x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 2.12 and 2.12. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x+13)(x 1)
= x^{4} + 22x^{3} + 131x^{2} + 158x 312 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x+13)(x 1) = x^{4} + 22x^{3} + 131x^{2} + 158x 312
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,1200), (11,840), (8,360), (5,48), (2,264), (1,0), (4,4080) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 13. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x+13)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 66x^{2} + 262x + 158
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 66x^{2} + 262x + 158 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.7, 5.0, 0.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 132x^{2} + 131x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.4 and 2.6. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x+13)(x 1)
= x^{4} + 14x^{3} 13x^{2} 314x + 312 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x+13)(x 1) = x^{4} + 14x^{3} 13x^{2} 314x + 312
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,2160), (11,1800), (8,1080), (5,432), (2,792), (1,0), (4,0), (7,4680) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 13. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x+13)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 42x^{2} 26x 314
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 42x^{2} 26x 314 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.3, 2.7, 2.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 84x^{2} 13x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 7.3 and 0.3. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x+13)(x 4)
= x^{4} + 19x^{3} + 62x^{2} 304x 1248 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x+13)(x 4) = x^{4} + 19x^{3} + 62x^{2} 304x 1248
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,1440), (11,1050), (8,480), (5,72), (2,528), (1,1470), (4,0), (7,8580) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 13. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x+13)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 57x^{2} + 124x 304
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 57x^{2} + 124x 304 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.6, 4.9, 1.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 114x^{2} + 62x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.25 and 1.25. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)^{2}(x+13)
= x^{4} + 11x^{3} 58x^{2} 320x + 1248 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)^{2}(x+13) = x^{4} + 11x^{3} 58x^{2} 320x + 1248
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,2592), (11,2250), (8,1440), (5,648), (2,1584), (1,882), (4,0), (7,2340) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a double root at x = 4. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 13. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)^{2}(x+13) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 33x^{2} 116x 320
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 33x^{2} 116x 320 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.3, 1.9, 4.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 66x^{2} 58x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 6.9 and 1.4. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x+13)(x+5)
= x^{4} + 28x^{3} + 269x^{2} + 1082x + 1560 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x+13)(x+5) = x^{4} + 28x^{3} + 269x^{2} + 1082x + 1560
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,720), (11,420), (8,120), (5,0), (2,264), (1,2940) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 13. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x+13)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 84x^{2} + 538x + 1082
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 84x^{2} + 538x + 1082 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [11.0, 5.5, 4.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 168x^{2} + 269x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 9.04 and 4.96. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x+13)(x+5)
= x^{4} + 20x^{3} + 77x^{2} 302x 1560 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x+13)(x+5) = x^{4} + 20x^{3} + 77x^{2} 302x 1560
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(14,1296), (11,900), (8,360), (5,0), (2,792), (1,1764), (4,0), (7,9360) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 13. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x+13)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 60x^{2} + 154x 302
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 60x^{2} + 154x 302 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [10.7, 5.4, 1.3]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 120x^{2} + 77x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 8.49 and 1.51. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 1)(x 4)
= x^{4} + 5x^{3} 22x^{2} 80x + 96 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 1)(x 4) = x^{4} + 5x^{3} 22x^{2} 80x + 96
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,264), (4,0), (1,150), (2,96), (5,396), (8,4704) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 1. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 1)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 15x^{2} 44x 80
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 15x^{2} 44x 80 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.1, 1.3, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 30x^{2} 22x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.54 and 1.04. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)^{2}(x 1)
= x^{4} 3x^{3} 30x^{2} + 128x 96 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)^{2}(x 1) = x^{4} 3x^{3} 30x^{2} + 128x 96
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,968), (4,640), (1,250), (2,32), (5,44), (8,1568) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a double root at x = 4. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)^{2}(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 9x^{2} 60x + 128
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 9x^{2} 60x + 128 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.8, 2.1, 4.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 18x^{2} 30x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 1.61 and 3.11. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 15)(x 1)
= x^{4} 6x^{3} 121x^{2} 234x + 360 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 15)(x 1) = x^{4} 6x^{3} 121x^{2} 234x + 360
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,528), (4,0), (1,480), (2,624), (5,3960), (8,8232), (11,10200), (14,4680), (17,15456), (20,59280) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 15. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 15)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 18x^{2} 242x 234
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 18x^{2} 242x 234 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.1, 1.0, 10.7]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 36x^{2} 121x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.23 and 6.23. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x 15)(x 1)
= x^{4} 14x^{3} 41x^{2} + 414x 360 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x 15)(x 1) = x^{4} 14x^{3} 41x^{2} + 414x 360
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,1936), (4,1520), (1,800), (2,208), (5,440), (8,2744), (11,4760), (14,2600), (17,9568), (20,39520) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 15. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x 15)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 42x^{2} 82x + 414
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 42x^{2} 82x + 414 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.5, 2.6, 11.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 84x^{2} 41x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.87 and 7.87. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 15)(x 4)
= x^{4} 9x^{3} 106x^{2} + 144x + 1440 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 15)(x 4) = x^{4} 9x^{3} 106x^{2} + 144x + 1440
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,726), (4,0), (1,1200), (2,1248), (5,990), (8,4704), (11,7140), (14,3600), (17,12558), (20,49920) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 15. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 15)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 27x^{2} 212x + 144
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 27x^{2} 212x + 144 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.0, 0.7, 11.2]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 54x^{2} 106x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 2.52 and 7.02. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)^{2}(x 15)
= x^{4} 17x^{3} 2x^{2} + 576x 1440 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)^{2}(x 15) = x^{4} 17x^{3} 2x^{2} + 576x 1440
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,2662), (4,2432), (1,2000), (2,416), (5,110), (8,1568), (11,3332), (14,2000), (17,7774), (20,33280) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a double root at x = 4. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 15. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)^{2}(x 15) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 51x^{2} 4x + 576
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 51x^{2} 4x + 576 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.0, 4.0, 11.9]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 102x^{2} 2x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.04 and 8.54. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 15)(x+5)
= x^{4} 151x^{2} 990x 1800 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 15)(x+5) = x^{4} 151x^{2} 990x 1800
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,132), (4,0), (1,960), (2,4368), (5,9900), (8,15288), (11,16320), (14,6840), (17,21252), (20,78000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 15. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 15)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 302x 990
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 302x 990 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 4.4, 10.1]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 151x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.02 and 5.02. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x 15)(x+5)
= x^{4} 8x^{3} 119x^{2} + 90x + 1800 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x 15)(x+5) = x^{4} 8x^{3} 119x^{2} + 90x + 1800
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,484), (4,304), (1,1600), (2,1456), (5,1100), (8,5096), (11,7616), (14,3800), (17,13156), (20,52000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 15. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x 15)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 24x^{2} 238x + 90
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 24x^{2} 238x + 90 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 0.4, 11.2]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 48x^{2} 119x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 2.88 and 6.88. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 1)(x+5)
= x^{4} + 14x^{3} + 59x^{2} + 46x 120 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 1)(x+5) = x^{4} + 14x^{3} + 59x^{2} + 46x 120
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,48), (4,0), (1,120), (2,336), (5,3960) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 1. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 1)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 42x^{2} + 118x + 46
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 42x^{2} + 118x + 46 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 4.4, 0.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 84x^{2} + 59x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.05 and 1.95. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x 1)(x+5)
= x^{4} + 6x^{3} 21x^{2} 106x + 120 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x 1)(x+5) = x^{4} + 6x^{3} 21x^{2} 106x + 120
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,176), (4,80), (1,200), (2,112), (5,440), (8,5096) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 1. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x 1)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 18x^{2} 42x 106
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 18x^{2} 42x 106 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 1.7, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 36x^{2} 21x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.9 and 0.9. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x+4)(x 1)
= x^{4} + 15x^{3} + 68x^{2} + 60x 144 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x+4)(x 1) = x^{4} + 15x^{3} + 68x^{2} + 60x 144
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,24), (4,0), (1,150), (2,384), (5,4356) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 4. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x+4)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 45x^{2} + 136x + 60
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 45x^{2} + 136x + 60 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 4.7, 0.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 90x^{2} + 68x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.4 and 2.1. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x 4)(x 1)
= x^{4} + 7x^{3} 20x^{2} 132x + 144 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x 4)(x 1) = x^{4} + 7x^{3} 20x^{2} 132x + 144
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,88), (4,160), (1,250), (2,128), (5,484), (8,5488) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 4. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x 4)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 21x^{2} 40x 132
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 21x^{2} 40x 132 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 2.0, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 42x^{2} 20x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 4.28 and 0.78. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x+4)(x 4)
= x^{4} + 12x^{3} + 20x^{2} 192x 576 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x+4)(x 4) = x^{4} + 12x^{3} + 20x^{2} 192x 576
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,33), (4,0), (1,375), (2,768), (5,1089), (8,9408) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 4. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x+4)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 36x^{2} + 40x 192
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 36x^{2} + 40x 192 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 4.7, 1.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 72x^{2} + 20x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.38 and 0.62. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x 4)^{2}
= x^{4} + 4x^{3} 44x^{2} 96x + 576 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x 4)^{2} = x^{4} + 4x^{3} 44x^{2} 96x + 576
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,121), (4,256), (1,625), (2,256), (5,121), (8,3136) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a double root at x = 4. Not just the function but also its first derivative are zero at this point. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x 4)^{2} = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 12x^{2} 88x 96
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 12x^{2} 88x 96 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 1.0, 4.0]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 24x^{2} 44x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.89 and 1.89. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x+4)(x+5)
= x^{4} + 21x^{3} + 164x^{2} + 564x + 720 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x+4)(x+5) = x^{4} + 21x^{3} + 164x^{2} + 564x + 720
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,6), (4,0), (1,300) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 4. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x+4)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 63x^{2} + 328x + 564
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 63x^{2} + 328x + 564 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 5.3, 4.3]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 126x^{2} + 164x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.73 and 4.77. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)^{2}(x 4)(x+5)
= x^{4} + 13x^{3} + 28x^{2} 204x 720 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)^{2}(x 4)(x+5) = x^{4} + 13x^{3} + 28x^{2} 204x 720
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,22), (4,32), (1,500), (2,896), (5,1210), (8,10192) Let us inspect the roots of the given polynomial function.
There is a double root at x = 6. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 4. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)^{2}(x 4)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 39x^{2} + 56x 204
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 39x^{2} + 56x 204 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [6.0, 5.3, 1.6]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 78x^{2} + 28x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.68 and 0.82. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 8)(x 1)
= x^{4} + 1x^{3} 58x^{2} 136x + 192 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 8)(x 1) = x^{4} + 1x^{3} 58x^{2} 136x + 192
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,360), (4,0), (1,270), (2,288), (5,1188), (8,0), (11,7650) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 8. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 8)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 3x^{2} 116x 136
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 3x^{2} 116x 136 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.1, 1.1, 5.6]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 6x^{2} 58x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.37 and 2.87. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x 8)(x 1)
= x^{4} 7x^{3} 34x^{2} + 232x 192 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x 8)(x 1) = x^{4} 7x^{3} 34x^{2} + 232x 192
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,1320), (4,960), (1,450), (2,96), (5,132), (8,0), (11,3570) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 8. There is a single, unique root at x = 1. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x 8)(x 1) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 21x^{2} 68x + 232
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 21x^{2} 68x + 232 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.6, 2.5, 6.5]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 42x^{2} 34x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 1.2 and 4.7. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 8)(x 4)
= x^{4} 2x^{3} 64x^{2} + 32x + 768 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 8)(x 4) = x^{4} 2x^{3} 64x^{2} + 32x + 768
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,495), (4,0), (1,675), (2,576), (5,297), (8,0), (11,5355) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 8. There is a single, unique root at x = 4. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 8)(x 4) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 6x^{2} 128x + 32
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 6x^{2} 128x + 32 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.0, 0.3, 6.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 12x^{2} 64x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 2.8 and 3.8. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)^{2}(x 8)
= x^{4} 10x^{3} 16x^{2} + 352x 768 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)^{2}(x 8) = x^{4} 10x^{3} 16x^{2} + 352x 768
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,1815), (4,1536), (1,1125), (2,192), (5,33), (8,0), (11,2499) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a double root at x = 4. Not just the function but also its first derivative are zero at this point. There is a single, unique root at x = 8. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)^{2}(x 8) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 30x^{2} 32x + 352
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 30x^{2} 32x + 352 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [3.2, 4.0, 6.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 60x^{2} 16x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.49 and 5.49. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x+4)(x 8)(x+5)
= x^{4} + 7x^{3} 46x^{2} 472x 960 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x 8)(x+5) = x^{4} + 7x^{3} 46x^{2} 472x 960
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,90), (4,0), (1,540), (2,2016), (5,2970), (8,0), (11,12240) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 8. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x+4)(x 8)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} + 21x^{2} 92x 472
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} + 21x^{2} 92x 472 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 4.4, 4.8]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} + 42x^{2} 46x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 5.03 and 1.53. At these points, the concavity changes.
Example:Let's analyze the following polynomial function. f(x) = (x+6)(x 4)(x 8)(x+5)
= x^{4} 1x^{3} 70x^{2} 8x + 960 (obtained on multiplying the terms)
You might also be interested in reading about quadratic and cubic functions and equations.
Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x 4)(x 8)(x+5) = x^{4} 1x^{3} 70x^{2} 8x + 960
We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(7,330), (4,192), (1,900), (2,672), (5,330), (8,0), (11,5712) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = 6. There is a single, unique root at x = 4. There is a single, unique root at x = 8. There is a single, unique root at x = 5. These roots are the solutions of the quartic equation f(x) = 0 These values of x are the roots of the quadratic equation (x+6)(x 4)(x 8)(x+5) = 0
Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x^{3} 3x^{2} 140x 8
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x^{3} 3x^{2} 140x 8 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [5.5, 0.0, 6.4]. At these points, the curve has either a local maxima or minima. These are the extrema  the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..
Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x^{3} 6x^{2} 70x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 3.17 and 3.67. At these points, the concavity changes.

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