# Applications of Differentiation – Extreme Value Theorem – Global Extrema

We have learned how to use the first derivative test to find critical numbers, local extrema, and intervals of increasing and decreasing.

In this lesson though, we will use the first derivative test and the Extreme Value Theorem to find the global extrema of a function.

Before we look at a function, let’s consider a mountain range like below: Over the span of this mountain range, there is a high point (H) and low point (L): As a matter of fact, for ANY stretch of land of a finite length L, there will be a high point and a low point.

Similarly, the graph of a function over an interval [a, b] of length L = b – a there will be a high point and low point, which is essentially what the Extreme Value Theorem explains.

 Definition: Extreme Value Theorem (EVT) If a function F(x) is continuous on a closed interval [ a, b], then F(x) has both a maximum and minimum value on the interval [a, b].

In other words, a continuous function on a closed interval must attain its highest or lowest value somewhere on the closed interval.

On a closed interval of any function, there are 2 different types of extrema:

• Local Maximum/Minimum: the small or the large value of the function in a small domain. These values will NOT occur on the endpoints. Rather, these values will be within the interval.
• Absolute Maximum/Minimum: The greatest or smallest value of the function over any given interval [a, b]. The greatest or smallest value of a function can occur within the interval or at the endpoints of the interval, (a, F(a)) or (b, F(b)). The figure below demonstrates an example of this: Consider the curve on the figure below: On the closed interval [a,e] we see that the function has a local maximum at x = b, a local minimum at x = c, a global maximum at x = d, and a global minimum at x = a.

Consider another curve below: On the closed interval [a,b] we see that the function has a global maximum at x = b, and a global minimum at x = c. Even though the function is not continuous at x = c, we still have a minimum value at that location.

There are 3 key points regarding the Extreme Value Theorem:

• Global extrema can occur within the interval or at the endpoints of the interval.
• Local extrema are ALWAYS located within the interval.
• A function can be discontinuous on the interval and still have a value at the location of discontinuity.
 Guidelines for Applying the Extreme Value Theorem Take the first derivative of the function. Set the first derivative equal to zero and solve for the variable to find the critical numbers of the function. Plot the critical numbers on a number line. Substitute values from the number line into the derivative to determine the intervals of increasing and decreasing. Substitute the critical numbers into the original function to obtain the y coordinate of the critical number. Substitute the value of a and b into the original function to obtain the y coordinate of a and b. You must test the endpoints at x = a and x = b because the global extrema can be at those locations. List the critical points and end points and determine the global extrema of the function on the interval.

Example 1:

Suppose we are given with the function value as, F(x) = x3 – 2x2 -1 on the given interval [-3, 3]. Find the following:

• Local extrema
• Global extrema

Solution:

First, Take the first derivative of the function.

F’ (x) = 3x2 – 4x

Second, set the first derivative equal to zero and solve for the variable to find the critical numbers of the function.

0 = 3x2 – 4x

Factor a y from each term:

0 = x (3x-4)

Set each factor equal to zero:

0 = x

3x – 4=0

Solve for y:

x = 0

x = 4/3

x = 0 and x = 4/3 are the critical numbers.

Third, plot the critical numbers on a number line. Substitute values from the number line to F’(x) to determine intervals of increasing and decreasing. Let’s find a value between -∞ and 0 to plug into F’(x), such as -1:

F’ (x) = 3x2 – 4x

F’ (-1) = 3(-1)2 – 4(-1)

F’ (-1) = 3(1) + 4

F’ (-1) = 7

Since the first derivative is positive, the function is increasing on the interval of (- ∞, 0).

Let’s find a value between 0 and 4/3 to plug into F’(x), such as 1:

F’ (x) = 3x2 – 4x

F’ (1) = 3(1)2 – 4(1)

F’ (1) = 3(1) – 4

F’ (1) = 3 – 4

F’ (1) = -1

Since the first derivative is negative, the function is decreasing on the interval of (0 , 4/3).

Let’s find a value between 4/3 and ∞ to plug into F’(y), such as 2:

F’ (x) = 3x2 – 4x

F’ (2) = 3(2)2 – 4(2)

F’ (2) = 3(4) – 8

F’ (2) = 12 – 8

F’ (2) = 4

Since the first derivative is positive, the function is increasing on the interval of (4/3, ∞). Looking at the number line, we see that at xc= 0, F(x) has a local maximum. At xc= 4/3, F(x) has a local minimum.

Fourth, substitute the critical numbers into the original function to obtain the y coordinate of the critical number.

Since xc= 0 and xc= 4/3, substitute them into the function:

For xc= 0,

F(x) = x3 – 2x2 -1

F(0) = 0 – 0 – 1

F(0) = -1

For xc= 4/3,

F(4/3) = x3 – 2x2 -1

F(4/3) = (4/3)3-2(4/3)2-1

F(4/3) = -2.185

Our two critical points are (0,-1), which is the local maximum, and (4/3, -2.185), which is the local minimum.

Fifth, substitute the value of a and b into the original function to obtain the y coordinate of a and b.

According to the problem, a = -3 and b = 3

Putting a = -3 inside F(x),

F (-3) = (-3)3 – 2(-3)2 -1

F (-3) = -27 -18-1

F (-3) = -46

At a = -3, we have the point (-3, -46)

Putting b = 3 inside F(x),

F (3) = (3)3 – 2(3)2 -1

F (3) = 27-18-1

F (3) = 8.

At b = 3, we have the point (3, 8).

Sixth, list the critical points and endpoints and determine the global extrema of the function on the interval.

Our critical points are:

• (0,-1)
• (4/3, 2.185)

Our endpoints are:

• (3,8)
• (-3,-46)

Hence the minimum value (global minimum) of the function will be y = -46 at x = -3 and maximum value (global maximum) of the function will be y = 8 at x = 3.

Let’s graph the function to verify our results: Example 2:

For F (x) = sin(x) on the interval [0 to 2π], find the following:

• Local Extrema
• Global Extrema

Solution:

First, take the first derivative of the function.

F’ (x) = cos(x)

Second, set the first derivative equal to zero and solve for the variable to find the critical numbers of the function.

0 = cos(x)

The x values that will make cos(x) equal to zero on the interval of [0, 2π] would be 2/π and 3π/2. This can be determined by looking at the unit circle: Therefore, x = 2/π and x =3π/2 are the critical numbers of the function.

Third, plot the critical numbers on a number line. Substitute values from the number line to F’(x) to determine intervals of increasing and decreasing. Let’s find a value between -∞ and π/2 to plug into F’(x), such as 0:

F’ (x) = cos(x)

F’ (0) = cos(0)

F’ (x) = 1

Since the first derivative is positive, the function is increasing on the interval of (- ∞, π/2).

Let’s find a value between 2/π and 3π/2 to plug into F’(x), such as π:

F’ (x) = cos(x)

F’ (π) = cos(π)

F’ (π) = -1

Since the first derivative is negative, the function is decreasing on the interval of

(2/π,3π/2).

Let’s find a value between 3π/2 and to plug into F’(y), such as 5π/3:

F’ (x) = cos(x)

F’ (5π/3) = cos(5π/3)

F’ (5π/3) = ½

Since the first derivative is positive, the function is increasing on the interval of (3π/2, ∞). Looking at the number line, we see that at xc= 2/π , F(x) has a local maximum. At xc= 3π/2 , F(x) has a local minimum.

Fourth, substitute the critical numbers into the original function to obtain the y coordinate of the critical number.

Since xc= 2/π and xc= 3π/2, substitute them into the function:

For xc= 2

F(2) = sin (2) = 1

For xc= 3π/2,

F(3π/2) = sin (3π/2) = -1

Our critical points are (2, 1), which is the local maximum, and (3π/2, -1), which is the local minimum.

Fifth, substitute the value of a and b into the original function to obtain the y coordinate of a and b.

According to the problem, a = 0 and b = 2

F(x) = sin(x)

F(0) = sin(0)

F(0) = 0

At a = 0, we have the point (0,0)

Putting b = 2π inside F(x),

F(x) = sin(x)

F(2π) = sin(2π)

F(2π) = 0

At b = 2π, we have the point (2π, 0).

Sixth, list the critical points and endpoints and determine the global extrema of the function on the interval.

Our critical points are:

• ( 2/π, 1)
• (3π/2, -1)

Our endpoints are:

• (0, 0)
• (2π, 0)

Hence the minimum value (global minimum) of the function will be y = -1 at x = 3π/2 and maximum value (global maximum) of the function will be y = 1 at x = 2/π. Coincidentally, these are also the local extrema of the function which we determined as our critical points.

Let’s graph the function to verify our results: Summary of Section Extreme Value Theorem: If a function is continuous on the closed interval [a, b], then the function must attain a maximum value and a minimum value. To find the extrema of a function, apply the first derivative test to find the critical numbers and substitute those numbers back into the original function to obtain their y values. Don’t forget to test the endpoints at x = a and x = b to see if these x values produce a maximum or minimum value.