Definition.Definition. A function f is said to have a relative maximum at x0 if there is an open interval containing x0 on which f(x0) is the largest value, that is f(x0) ≥ f(x) for all x in the interval.

A function f is said to have a relative minimum at x0 if there is an open interval containing x0 on which f(x0) is the smallest value, that is f(x0) ≤ f(x) for all x in the interval.

If f has either a relative maximum or relative minimum at x0 , we say that it has a relative extremum at x0.

Analysis of Functions IIRelative Extrema

First and Second Derivative Tests

y = f(x).

relative maxima

relative minima

Example. Look at the three graphs below. The first has a relative minimum at 0 and no relative maxima. The second has a relative maximum at and a relative minimum at . The third has relative minima at plus or minus multiples of 2πand relative maxima at plus or minus multiples of 2π.

(1) y = x4

(2) y = − x3+ 4 x+4

(3) sin(x)

23

23

−

2π−

2π

Theorem. If a function f has any relative extrema, then they occur either at a point where f′ (x) = 0 or a point where the derivative of f does not exist.

Proof. Suppose that f has a derivative at the point a, and has a relative maximum there. Then if h < 0,

( ) ( ) 0f a h f ah

+ − >

Since this is true for all h < 0, we must have

( ) ( ) ( ) ( )lim lim 00 0

f a h f a f a h f ah hh h

+ − + −= ≥−→ →

If h > 0, ( ) ( ) 0f a h f ah

+ − <

Since this is true for all h > 0, we must have

( ) ( ) ( ) ( )lim lim 00 0

f a h f a f a h f ah hh h

+ − + −= ≤+→ →

Therefore we have ( ) ( )lim 0.0

f a h f ahh

+ − =→

A similar result holds at a relative minimum.

Positive slopeNegative slope

Definition. A critical point of f is any point where f does not have a derivative, or where the derivative of f is 0. If we only want to refer to points where the derivative is 0, we call them stationary points.

The previous result shows that all relative extreme points are critical points, so if we are looking for relative extreme points, we can find critical points (a much easier job) and test each one. However, not all critical points are relative extreme points as the following table of figures shows.

Critical points that are relative extrema. The derivative has opposite signs on either side of the critical point.

Critical points that are not relative extrema. The derivative has the same sign on either side of the critical point.

Theorem. (First Derivative Test) Suppose that f is continuous at a critical point x0 and has a derivative in some interval containing x0, except possibly at x0.

(a) If f′ (x) > 0 in an interval to the left of x0, and f′ (x) < 0 in an open interval to the right of x0, then f has a relative maximum at x0.

(b) If f′ (x) < 0 in an interval to the left of x0, and f′ (x) > 0 in an open interval to the right of x0, then f has a relative minimum at x0.

(c) (a) If f′ (x) has the same sign at all points of an open interval containing x0, except at x0, then f does not have a relative extreme point at x0.

Solution. Since f is a polynomial it has a derivative everywhere, so all critical points are stationary points.

2 2( ) 3 6 3 3( 1)f x x x x′ = + + = +

Thus the only stationary point is at x = −1. If x is not equal to −1, then the derivative is always positive. Thus the point is nota relative extreme point.

Example. Let

Find all critical points and use the first derivative test to see if they are extreme points.

3 2( ) 3 3 1f x x x x= + + +

Here is the graph.

Solution.

This function is 0 at x = 1, and becomes infinite at x = 0, so that f does not have a derivative at 0. Therefore the critical points are 0 and 1.

1 2 23 3 3

23

( 1)( ) ( 1) xf x x x x x

x

− − −′ = − = − =

If x < 0, (x – 1) < 0, and if 0 < x< 1, (x – 1) < 0. On the other hand the denominator is always positive, so f′ (x) is negative on either side of 0, and 0 is not a relative extreme point.

Example. Let

Find all critical points and use the first derivative test to see if they are extreme points.

4 13 3 34

( ) 3f x x x= −

If x <1, (x – 1) < 0, and if x> 1, (x – 1) > 0. Thus f′ (x) is negative on the left of 1, and positive on the right. This meansthat 1 is a relative minimum of f. The graph is shown below.

There is another way to test for extreme points, using shape rather than direction, that is using the second derivative.

Theorem. (the second derivative test) Suppose that f is twice differentiable at x0. Then

(a) If f′ (x0) = 0 and f′′ (x0) > 0, then f has a relative minimum at x0.

(b) If f′ (x0) = 0 and f′′ (x0) < 0, then f has a relative maximum at x0.

(c) If f′ (x0) = 0 and f′′ (x0) = 0, then the test is inconclusive.

+ + +

− − −

f′ (x0) = 0 and f′′ (x0) > 0 f′ (x0) = 0 and f′′ (x0) < 0

Example. Find all relative extrema ofand identify them.

4 2( ) 2f x x x= −

Solution. 3 2( ) 4 4 4 ( 1) 4 ( 1)( 1)f x x x x x x x x′ = − = − = − +Thus the critical points are x = 0, x = 1, x = −1 (all stationary points). The second derivative is

2( ) 12 4f x x′′ = −

Thus (1) 8f ′′ =( 1) 12 4 8f ′′ − = − = (0) 4f ′′ =−

We see that 1 and –1 are local minima, and 0 is a local maximum.

The graph confirms these results.

Comparison of the two types of tests.

The second derivative test is certainly easiest to use when it works, but it has two problems.

First, it does not apply at a point where the derivative does not exist (since you cannot take the second derivative in that case), or at a point where the function has only one derivative.

Secondly, it may fail to give an answer because the second derivative may be zero.

Thus the strategy is to use the second derivative test if possible, but if it fails or cannot be used, use the first derivative test as a backup.

Solution 1. ( ) 2 6 2( 3)f x x x′ = − = −

The only critical point is x = 3. If x <3, then the derivative is negative, and if x > 3 the derivative is positive. Thus x = 3 is a relative minimum.

( ) 2 6 2( 3)f x x x′ = − = −Solution 2. and ( ) 2.f x′′ =

Since the second derivative is positive at x = 3, that must be a relative minimum.

Example. Use the first and second derivative tests to show that the function has a relative minimum at x = 3.2( ) 6 5f x x x= − +

Note. A relative maximum or minimum is sometimes called a local maximum or minimum.

This is confirmed by the graph.

Solution 1.

Solution 2.

Since and

2 2( ) 3 12 9 3( 4 3) 3( 3)( 1)f x x x x x x x′ = − + = − + = − −

2 2( ) 3 12 9 3( 4 3) 3( 3)( 1)f x x x x x x x′ = − + = − + = − −( ) 6 12 6( 2)f x x x′′ = − = −The critical points are x = 1 and x = 3.

(1) 6f ′′ =− (3) 18 12 6.f ′′ = − = Thus x = 1 is a localmaximum and x = 3 is a local minimum.

Example. Use the first and second derivative tests to show that the function has a local maximum at x = 1, and a local minimum at x = 3.

3 2( ) 6 9 1f x x x x= − + −

The critical points are x = 1 and x = 3. If x < 1, both factors are negative, and so f′ is positive. Between 0 and 1 one factor is positive and one negative, so f′ is negative. Finally, if x > 3, both factors are positive and f′ is positive. Since the derivative is positive to the left of 1 and negative to the right, x = 1 is a local maximum. Since the opposite is true at x = 3, this point is a local minimum.

Again, we can confirm this with the graph.

Solution. 4( ) 5( 2)f x x′ = − 3( ) 20( 2)f x x′′ = −

Since the derivative is 0 at x = 2 and exists everywhere else, x = 0 is the only critical point. But so the second derivative test is inconclusive. We must turn to the first derivative test, which always works.

(2) 0f ′′ =

Since the first derivative is always positive, x = 0 is not a local extreme point.

Example. Verify that the function has a stationary point at x = 2. What can we learn about this point using the first and second derivative tests?

5( ) ( 2)f x x= −

The graph is:

The graphs are as follows:

2 2( ) (2 2 ) xf x x x e−′ = − 2 2( ) 2 (1 4 2 )xf x e x x−′′ = − +

( )f x′ ( )f x′′

Example. Plot the graphs of the first and second derivative of the function on a graphing utility and use those results to estimate the x coordinates and type of the relative extreme points of f. Check with the graph of f.

2 2( ) xf x x e−=

It seems clear that x = 1 is a stationary point, and by the second graph it is a local maximum. Also, x = 0 seems to be a stationary point, and by the second graph, it is a local minimum.

Example. Find and classify the relative extrema of the function2

( )2 1

xf xx

=+

Solution.

( ) ( )2 2( 1)2 (2 ) 2( )

2 22 21 1

x x x x xf x

x x

+ −′ = =

+ +

The derivative exists everywhere and is zero at x = 0.

( ) ( )( )

( )( )( )

22 2 2 21 (2) 2 (2) 1 (2 ) 1 2 8 2( )

4 42 21 1

x x x x x x xf x

x x

+ − + + − +′′ = =

+ +

(0) 2f ′′ = so 0 is a local minimum.

Here is the graph.