Extreme Value Theorem - Formula, Examples, Proof, Statement

Extreme value theorem proves the existence of the maximum and minimum value of the function if the function is continuous in the closed interval [a, b]. In this article, we will discuss the extreme value theorem in depth along with the extreme value theorem statement, extreme value theorem proof and how to use the extreme value theorem. We will also solve some examples related to the extreme value theorem.

Let's start our learning on the topic "Extreme Value Theorem."

What is Extreme Value Theorem?

Extreme value theorem gives the maximum and minimum value of a real and continuous function in the specified closed interval. This theorem bounds the maximum and minimum value of the function with the closed interval. The Extreme value theorem is used to prove other theorems like Rolle's.

Extreme Value Theorem Statement

Extreme value theorem states that:

If a real-value function f is continuous at closed interval [p, q] then, there exists two points r and s in the given interval which gives the maximum value f(r) and minimum value f(s) for the given function f(x).

The graph for extreme value theorem is added below:

Extreme Value Theorem Formula

Extreme value theorem formula is given by:

f(s) ≤ f(x) ≤ f(r) ∀ x ∈ [p, q]

Extreme Value Theorem Proof

To prove the extreme value theorem, we use the contradiction and boundedness theorem. We will prove the function f has its maximum value in the interval [a, b]. Similarly, the function f has its minimum value in the interval [a, b] can be proved.

f is continuous on [a, b] means f is bounded on [a, b] such that there exists m and M i.e., m ≤ f(x) ≤ M by the Boundedness theorem.

Let M be the least upper bound of f(x) and at point, c lies in [a, b] i.e., f(c) = M, f(x) attains its maximum value on the interval [a, b]. Hence proved.

Let there is no such value c in [a, b] then, f(x) < M ∀x in [a, b].

A function g(x) is defined as g(x) = 1 / [M - f(x)] on [a, b]. We know that g(x) > 0 as f(x) < M ∀x in [a, b] and g is also continuous on [a, b].

By the boundedness theorem g(x) is also bounded on [a, b] which implies there exists k > 0 such that g(x) ≤ k, ∀x in [a, b].

⇒ 1 / [M - f(x)] ≤ k

⇒ M - f(x) ≥ 1 / k

Adding f(x) - (1 / k) on both sides of the inequality

⇒ M - (1/k) ≥ f(x)

⇒ f(x) ≤ M - (1/k)

The above expression contradicts that M is the least upper bound of f(x). Therefore, the assumption that there exists no such c in [a, b] so that f(c) = M is incorrect. Hence, f has its maximum on [a, b].

How to Use Extreme Value Theorem

Below are the steps to use the extreme value theorem.

Step 1: Check whether the function is differentiable or not.

Step 2: If a function is differentiable, it is a continuous function.

Step 3: Then, find the first derivative of the function and equate it to 0.

Step 4: Find the critical points of the function.

Step 5: Put the critical points in the function and find its value.

Step 6: Put the points present in the specified closed interval and find the function value.

Step 7: Identify the maximum and minimum value of the function from the obtained values.

• Minimum of the function is the minimum among obtained values and the point is called the point of minimum.
• Maximum of the function is the maximum among obtained values and the point is called the point of maximum.

Article Related to Extreme Value Theorem:

• Calculus in Maths
• Differential Calculus
• Integral Calculus
• Mean Value Theorem
• Cauchy's Mean Value Theorem

Examples on Extreme Value Theorem

Example 1: Find the extreme values of the function f(x) = x³ - 4x² + 4x + 6 in the interval [-1, 5].

Solution:

f(x) = x³ - 4x² + 4x + 6

First, find f'(x) and equate it to zero.

f'(x) = 3x² - 8x + 4

3x² - 8x + 4 = 0

x = 2, 2/3

The critical points of function are 2, 2/3.

Now, find the value of f(x) by putting critical points and range of interval.

f (2) = 2³ - 4(2)² + 4(2) + 6 = 6

f (2/3) = (2/3)³ - 4(2/3)² + 4(2/3) + 6 = 7.185

f (-1) = (-1)³ - 4(-1)² + 4(-1) + 6 = -3

f (5) = 5³ - 4(5)² + 4(5) + 6 = 51

So, the maximum value of the function is 51 at x = 5 and the minimum value of function is -3 at x = -1.

Example 2: Find the extreme values of the function p(x) = x² - 8x + 6 in the interval [2, 5].

Solution:

p(x) = x² - 8x + 6

First find p'(x) and equate it to 0

p'(x) = 2x - 8

2x - 8 = 0

2x = 8

x = 4

The critical point of the function p(x) is 4.

Now, find the value of the function p(x) at critical points and the range of interval.

p(4) = 4² - 8(4) + 6 = -10

p(2) = 2² - 8(2) + 6 = -9

p(5) = 5² - 8(5) + 6 = -6

Extreme values are -6 (maximum at x = 5) and -10 (minimum at x = 4).

Example 3: Find the extreme value of the function g(x) = 2sin x - 1 in the interval [0, 2π/3]

Solution:

g(x) = 2sin x - 1

First find g'(x) and equate it 0.

g'(x) = 2cos x

2cos x = 0

cos x = 0

x = π / 2

Now, find the values of the function at critical point and range of the interval.

g(π / 2) = 2sin ( π / 2) - 1 = 1

g(0) = 2 sin(0) - 1 = -1

g(2π/3) = 2sin(2π/3) - 1 = √3 - 1

Extreme values of the function g(x) are -1 (minimum x = 0) and 1 (maximum at x = π / 2)

Example 4: Find the extreme values of the function f(x) = x + cos x in the interval [0, π]

Solution:

f(x) = x + cos x

first find f'(x) and equate it to 0

f'(x) = 1 - sin x

1 - sin x = 0

sin x = 1

x = π / 2

The critical point of the function f(x) is π/2

Now, find the values of the function at critical point and range of interval'

f(π/2) = (π/2) + cos (π/2) = π/2

f(0) = 0 + cos 0 = 1

f(π) = π + cos π = π - 1

Extreme values of the function f(x) are 1 (minimum at x = 0) and π - 1 (maximum at x = π).

Practice Questions on Extreme Value Theorem

Q1. Find the extreme values of the function f(x) = 2x³ - 9x² + 2x + 10 in the interval [1, 6].

Q2. Find the extreme values of the function t(x) = x² - 9x + 2 in the interval [1, 4].

Q3. Find the extreme values of the function r(x) = sin x + cos x in the interval [0, π/2].

Q4. Find the extreme values of the function s(x) = x⁴ - 5x³ + x² - 2x -3 in the interval [0, 7].