Absolute Maxima and Minima

Definition. A function f is said to have an absolute maximum on an interval I at the point x0 if it is the largest value of f on that interval; that is if for all x in I.

A function f is said to have an absolute minimum on an interval I at the point x0 if it is the smallest value of f on that interval; that is if for all x in I.

If f has either an absolute maximum or an absolute minimum on an interval I at the point x0, we say that it has an absolute extremum at that point.

( ) ( )0

f x f x≥

( ) ( )0

f x f x≤

Absolute minimum on the interval [-8, 8]

Absolute maximum on the interval [-8, 8]

1. The function f was continuous on a closed interval (a finite interval and its end points), and it had an absolute maximum and an absolute minimum.

2. The absolute maximum and minimum occurred either at an end point or at a local maximum or minimum point.

Two facts appear in the previous example.

The two properties of f in part 1. above turn out to be critical. In general the existence of absolute maxima and minima will depend on the nature of the function f and the type of interval. The previous example demonstrates the following theorem.

Theorem. If f(x) is a continuous function on a closed interval [a, b], then f always has an absolute maximum and an absolute minimum value on [a, b].

This is a difficult theorem to prove, but it is intuitively clear. It says intuitively that if you put a pencil at f(a), and then draw a curve ending at f(b), without lifting the pencil from the paper, then that curve will have a highest and a lowest height.

The previous example also illustrates the proper procedure to find the absolute maximum and minimum values in this case.

To Find the Absolute maximum and minimum value of a continuous function f on a closed interval [a, b].

1. Locate all critical points in the interval [a, b]

2. Evaluate f at all of the critical points and at the points a and b.

3. The largest value found in step 2 is the absolute maximum of f on [a, b] and the corresponding point is the point where that maximum is achieved. A similar statement is true for the absolute minimum.

Example. Find the absolute extrema of In the interval [−1, 1].

4 13 3( ) 3 9f x x x= −

Solution.1 2 23 3 3

23

(4 3)( ) 4 3 (4 3) xf x x x x x

x

− − −′ = − = − =

The derivative of f does not exist at x = 0, and is 0 at the point x = 3/4 .Thus we must consider the four points −1, 0, ¾, 1. The values of the function at these four points are respectively:

( 1) 3 9 12f − = + = (0) 0f = 3 6.1334

f =−

(1) 3 9 6f = − =−

Thus the absolute maximum is 12 and it occurs at the left hand endpoint, and the absolute minimum is –6.133, occurring at x= ¾.

Solution.This function is 0 at the point x =

differs from this point by a multiple of π. Thus x =

( 1) 1.38f − = 1.4144

f π − =

(1) 0.301f =−

Thus the absolute maximum is 1.414 and it occurs at − π/4,and the absolute minimum is –0.301, occurring at x = 1.

( ) cos( ) sin( )f x x x′ = +

4π− and at every point that

4π−

is the only such point within the desired interval. We must therefore consider the points −1, − π/4, and 1.

Example. Find the absolute extrema of in the interval [−1, 1].

( ) sin( ) cos( )f x x x= −

This is the graph of that function.

Example. Find the absolute extrema of In the interval [0, 6].

2( ) 8f x x x= −

Solution.

This function is 0 at the point x = 4. Thus we must check the three points 0, 4, 6. At these points the value of f is resp.

( ) 8 2f x x′ = −

(0) 0f = (4) 32 16 16f = − = (6) 48 36 12f = − =

Thus the absolute maximum is 16 and occurs at x = 4, and the absolute minimum is 0, occurring at x = 0.

Here is the graph.

Example. Find the absolute extrema of In the interval [−2, 1].

3 2( ) 2 3 12f x x x x= − −

Solution.

The derivative of f is 0 at the points x = 2 and x = −1. However, only one of these points is in the desired interval, namely x = −1. Thus we must check the three points −2, −1, 1. At these points the value of f is resp.

Thus the absolute maximum is 7 and occurs at x = −1, and the absolute minimum is −13, occurring at x = 1.

2 2( ) 6 6 12 6( 2) 6( 2)( 1)f x x x x x x x′ = − − = − − = − +

( 2) 2( 8) 3(4) 12( 2) 4f − = − − − − =− ( 1) 2 3 12 7f − =− − + =

(1) 2 3 12 13f = − − =−

Function in the interval[− 2, 3]

Function in the interval[− 2, 1]

Example. Find the absolute extrema of In the interval [−3, 3].

( ) 6 4f x x= −

Solution.

The derivative of f is never 0, since it is equal to 4 if x < 3/2, and to −4 when x > 3/2. The derivative fails to exist at 3/2, so this is a critical point in the desired interval. We must therefore test at the three points −3, 3/2, and 3.

Thus the absolute maximum is 18 and it occurs at x = −3, and the absolute minimum is 0, occurring at x = 3/2.

( 3) 6 12 18f − = + = 32

6 6 0f

= − = (3) 6 12 6f = − =

Here is the graph.

Absolute maxima and minima on infinite or non closed intervals

If an interval is not open or not finite, then there may be no absolute maximum or minimum. The following diagrams illustrate this problem.

f(x) = 1/x has no absolute maximum or minimum in (0, 1). It has an absolute minimum of 1 in (0, 1], but no absolute maximum.

f(x) = tan(x) has no absolute maximum or minimum in the open interval (−π/2, π/2).

x = −π/2 x = π/2

has no absolute minimum in the open interval1

( ) 21f x

x=

+

(− ∞, ∞), but has an absolute maximum at x = 0.

has no absolute maximum or minimum in the 3( ) f x x=interval (− ∞, ∞).

has no absolute maximum in the interval (−∞, ∞),2( ) f x x=

but has an absolute minimum at x = 0. It also has no absolute maximum in the interval (−1, 1).

There are, however, conditions where something can be said.

Theorem. Suppose that a function f is continuous on an interval I (of any kind) and has exactly one relative extremum in that interval at a point x0. If the point x0 is a relative maximum, then it is an absolute maximum. If it is a relative minimum, then it is an absolute minimum.

We will not try to prove this, but instead will illustrate it with a diagram.

Relative maximum If it is eventually higher, it must pass through a relative minimum.

Example. Let

Find all absolute extrema (if any) for f on the interval (0, ∞).

3 2( ) 3 5f x x x= − +

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

This function is zero at 0 and 2, and only x = 2 is in the given interval. This is the only critical point of the function in (0, ∞).

Since ( ) 6 6 6( 1)f x x x′′ = − = − and this is greater than 0 at 2, weknow that there is a relative minimum at x = 2, and by the theorem, this must be an absolute minimum. There is no absolute maximum since

lim ( )f xx

=+∞→∞

This is confirmed by the graph.

Example. Let

Find all absolute extrema (if any) for f on the interval (− ∞, ∞).

2( ) 2(1 )

xf xx

=+

Solution.

Which is zero only at x = 0. The derivative is positive for x > 0 and negative for x < 0, so by the first derivative test, 0 is a local minimum. Since this is the only critical point in the interval, it is an absolute minimum.

We also see that the function is always between 0 and 1 and

lim ( ) 1.f xx

=→±∞

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

x x x x xf xx x

+ −′ = =+ +

The function never reaches 1, so there is no

absolute maximum.

The graph confirms this.

Example. Let x(t) = t − 3sin(t) and y(t) =4 − 3cos(t), for t in the interval [0, 10]. What are the highest and lowest points on this curve?

For curves defined by parametric equations, where the parametric interval is a closed interval [a, b] also have absolute maxima and minima, in fact their graph is contained in a finite rectangle.

We know that dy

dy dtdxdxdt

= and so the stationary points of the

curve occur where 0.dydt

= Thus absolute maxima or

minima must occur either at the end points of the parametric interval or at these stationary points.

In the previous example, 3sin( ) 0dy tdt

= = at ,π 2 ,π and 3 .π

If we check these points and 0 and 10, we have values1, 7, 1, 7, 6.517, resp. Thus the absolute maximum of 7 is reached at points ,π and 3 .π The absolute minimum of 1is reached at 0 and 2 .π