Applications of differentiation - the graph of a function and its derivatives
Extreme points, local (or relative) maximum and local minimum
The first derivative test
The second derivative test and concavity
Finding and classifying critical (or stationary) points
Finding extreme points
Points of inflection
Finding points of inflection
Finding and classifying critical (or stationary) points, example
Extreme points, local (or relative) maximum and local minimum
The derivative  f '(x0) shows the rate of change of the function with respect to the variable x at the point x0.
The derivative  f '(x) is equal to the slope of the tangent line at x. If the derivative is positive, the function is increasing at that point. If it is negative, the function is decreasing.
A point on the graph of a function at which its first derivative is zero, so that the tangent line is parallel to the x-axis, is called the stationary point or critical point.
A point at which a function attains its maximum value among all points where it is defined is called a global (or absolute) maximum. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum.
The first derivative test
Let  f (x) be a function and x = c a critical point of  f.
Test for local maximum.  If in some neighborhood of c the derivative  f '(x) is positive to the left of c and negative to the right, then c is a local (or relative) maximum of  f.
Test for local minimum.  If in some neighborhood of c the derivative  f '(x) is negative to the left of c and positive to the right, then c is a local (or relative) minimum of  f.
In the figure below, xmax denotes a local maximum and xmin a local minimum. The change of  f '(x) from positive to negative to positive is shown by tangents.
A stationary (or critical) point at which the first derivative of a function changes sign, so that its graph does not cross a tangent parallel to the x-axis, is called the turning point.
If the second derivative is negative ( f ''(x) < 0 ) at a stationary point, it is local maximum of the function, and if the second derivative is positive ( f ''(x) > 0 ), it is local minimum, as shows the above figure.
The second derivative test and concavity
A function is said to be concave down on an interval if its first derivative is decreasing on the interval.
At the same time the second derivative of the function is negative on the interval.
A function is said to be concave up on an interval if its first derivative is increasing on the interval.
At the same time the second derivative of the function is positive on the interval.
The second derivative test shows when a critical point c is a local maximum or local minimum.
(1)  If  f '(c) equals 0 and  f ''(c) < 0, then  x = c is a local maximum.
(2)  If  f '(c) equals 0 and  f ''(c) > 0, then  x = c is a local minimum.
(3)  If  f '(c) equals 0 and  f ''(c) = 0, then c may be a local maximum, local minimum or point of inflection.
Thus, the second derivative is inconclusive when  f ''(c) = 0.
Finding and classifying critical or stationary points
Therefore, to find extreme points of a differentiable function y f (x) calculate its derivative, equate it to zero and solve for x.
Roots of the equation  f '(x) = 0 are potential abscissas of, maximums, minimums and points of inflections.
So, to classify these points we plug each root x0, x1, x2, . . . into successive higher order derivatives, starting with the second derivative.
If first of higher order derivatives that do not vanishes (becomes 0) at this point (say x0) is of even order, then the function has either maximum or minimum at that point, precisely
- a local maximum if  f (2n)(x0) < 0  or   a local minimum if  f (2n)(x0) > 0
If first of the higher order derivatives that do not vanishes at this point is of odd order, then the function has not extreme points (extremal points or extrema) at that point at all.
Points of inflection
A point of the graph of a function at which it crosses its tangent, and concavity changes from up to down or vice versa is called the point of inflection.
The second derivative of the function is zero and changes its sign at the point of inflection, as can be seen in the figure above.
Therefore, to find points of inflection of a differentiable function y f (x) calculate its second derivative, equate it to zero and solve for x.
A root of the equation  f ''(x) = 0 is the abscissa of a point of inflection if first of the higher order derivatives that do not vanishes at this point is of odd order.
Finding and classifying critical or stationary points example
Example:  Find extreme points and points of inflection of the quintic  y = -x5 + 5x4 - 7x3 + x2 + 4x -and sketch its graph.
Solution:  To rewrite given polynomial from general into translatable form, and this way determine its source form, we first calculate the coordinates of translation
then, plug  x0 and  y0  into
y + y0 = a5(x + x0)5 + a4(x + x0)4 + a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
to get the source form of the given polynomial
y + 1 = - 1(x + 1)5 + 5(x + 1)4 - 7(x + 1)3 + (x + 1)2 + 4(x + 1) - 1,
what after expanding and reducing gives   y = - x5 + 3x3,    the source quintic.
Therefore, we can write   y = -x5 + 5x4 - 7x3 + x2 + 4x -1
or     y - 1 = - (x - 1)5 + 3(x - 1)3    or     f (x) = - (x - 1)5 + 3(x - 1)3 + 1.
To find the extreme points, calculate the first derivative of the above expression set it to zero and solve for x
f ' (x) - 5(x - 1)4 + 9(x - 1)2 - (x - 1)2 · [5(x - 1)2 - 9]
then we set         f ' (x) = 0,       - (x - 1)2 · [5(x - 1)2 - 9] = 0
what yields,  x1 = 1x2 = 1 - 3/Ö5  and  x3 = 1 + 3/Ö5  as potential abscissas of the extreme points.
To check if  x1 = 1 is the abscissa of the extreme point, plug it into successive higher order derivatives, starting with the second derivative. Thus,
f '' (x) - 20(x - 1)3 + 18(x - 1) - 2(x - 1) · [10(x - 1)2 - 9]   since    f '' (1) = 0
we must proceed with the higher order derivatives,
f ''' (x) =   - 60(x - 1)2 + 18   so that    f ''' (1) = 18 is not 0
what classifies root  x1 = 1 as the abscissa of the point of inflection, as first of higher order derivatives that do not vanishes at this point is of odd order.
While roots,  x2 = 1 - 3/Ö5  and  x3 = 1 + 3/Ö5  plugged into the second derivative respectively yield
f '' (1 - 3/Ö5) > 0,    so  x2 = 1 - 3/Ö5   is the abscissa of the minimum
and       f '' (1 + 3/Ö5) < 0,    so  x3 = 1 + 3/Ö5   is the abscissa of the maximum.
To find points of inflection of the given quintic we set the second derivative to zero and solve for x
f '' (x) - 20(x - 1)3 + 18(x - 1),     - 2(x - 1) · [10(x - 1)2 - 9]  = 0
what yields, x1 = 1x2 = 1 - 3/Ö10  and  x3 = 1 + 3/Ö10 as possible abscissas of the points of inflection.
The root  x1 = 1 we've already classified as the point of inflection.
Then we plug roots,  x2 = 1 - 3/Ö10  and  x3 = 1 + 3/Ö10  into the third derivative respectively
f ''' (x) =   - 60(x - 1)2 + 18,   so     f ''' (1 - 3/Ö10) ¹ 0   and     f ''' ( 1 + 3/Ö10) is not 0
what classifies roots,  x2  and  xas the abscissas of the points of inflection.
By plugging the abscissas x1, x2 and x3 into the given quintic we get the ordinates of the points of inflection.
The graph of the given quintic polynomial and its source function is shown in the figure below.
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