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Extreme Value Theorem
The extreme value theorem is an important theorem in calculus that is used to find the maximum and minimum values of a continuous realvalued function in a closed interval. This theorem is used to prove Rolle's theorem in calculus. The extreme value theorem is specific as compared to the boundedness theorem which gives the bounds of the continuous function on a closed interval.
In this article, we will discuss the concept of extreme value theorem, its statement, and its proof. We will also learn how to use the theorem with the help of a few solved examples for a better understanding of the concept.
1.  What is Extreme Value of Theorem? 
2.  Extreme Value Theorem Statement 
3.  Extreme Value Theorem Proof 
4.  How to Use Extreme Value Theorem? 
5.  FAQs on Extreme Value Theorem 
What is Extreme Value of Theorem?
The extreme value theorem helps in proving the existence of the maximum and minimum values of a realvalued continuous function over a closed interval. Once the existence of maximum and minimum values is proved, we might be asked to determine those values using the derivative of the function and finding the critical points. Rolle's theorem and Mean value theorem are the consequences of the extreme value theorem. Let us understand the meaning of extreme value below as we proceed to state the theorem and prove it.
Extreme Value Meaning
Extreme values of a function f(x) are the values y = f(x) which a function attains for a specific input x such that no other value of f(x) in the range is greater or less than these values. We have two types of extreme values: maximum and minimum. The maximum value of a function is a value such that no other value of the function can be greater than this and the minimum value of a function is a value such that no other value of the function is less than this value.
Extreme Value Theorem Statement
The extreme value theorem states that 'If a realvalued function f is continuous on a closed interval [a, b] (with a < b), then there exist two real numbers c and d in [a, b] such that f(c) is the minimum and f(d) is the maximum value of f(x). Mathematically, we can write the formula for the extreme value theorem as, f(c) ≤ f(x) ≤ f(d), ∀ x ∈ [a, b].
The extreme value theorem can also be stated as 'If a realvalued function f is continuous on [a, b], then f attains its maximum and minimum of [a, b].
Extreme Value Theorem Proof
Now that we have understood the extreme value theorem and its statement, let us now prove it using the contradiction method and the boundedness theorem. We will prove that f attains its maximum on the closed interval [a, b]. The proof that f attains its minimum on [a, b] can be proved on similar lines.
By hypothesis, f is continuous on [a, b], so f is bounded on [a, b] such that there exist m, M such that we have m ≤ f(x) ≤ M using the Boundedness Theorem. Here, suppose M is the least upper bound of f. Now, if there exists c in [a, b] such that f(c) = M, then this implies f attains maximum on [a, b]. We have proved the required result.
Now, assume there is no such c in [a, b], then we have f(x) < M for all x in [a, b]. Define a function h(x) = 1 / [M  f(x)] on [a, b]. Now, we know that h(x) > 0 because f(x) < M for all x in [a, b] and h is also continuous on [a, b]. So, using the boundedness theorem, we have h(x) is bounded on [a, b]. This implies there exists K > 0 such that h(x) ≤ K, for all x in [a, b].
⇒ 1 / [M  f(x)] ≤ K
⇒ M  f(x) ≥ 1/K
Adding f(x)  1/K on both sides, we have
⇒ M  1/K ≥ f(x)
⇒ f(x) ≤ M  1/K
This contradicts the fact that M is the least upper bound of f(x). Hence, our assumption that there exists no such c in [a, b] such that f(c) = M is wrong. Therefore, f attains its maximum on [a, b].
We can prove that f attains its minimum on [a, b] on similar lines.
How to Use Extreme Value Theorem?
Now that we have proved the extreme value theorem, let us learn how to use it with the help of an example. Consider function f(x) = x^{3}  27x + 2. Find the maximum and minimum values of f(x) on [0, 4] using the extreme value theorem.
Solution: Since f(x) = x^{3}  27x + 2 is differentiable, therefore it is continuous. Since [0, 4] is closed and bounded, therefore we can apply the extreme value theorem. Differentiate f(x) = x^{3}  27x + 2.
f'(x) = 3x^{2}  27
Setting f'(x) = 0, we have
3x^{2}  27 = 0
⇒ 3x^{2} = 27
⇒ x^{2} = 27/3 = 9
⇒ x = 3, 3
So, x = 3, 3 are the critical points. Now, we find the value of f(x) at critical points and the endpoints of the interval.
f(3) = (3)^{3}  27(3) + 2 = 27 + 81 + 2 = 56
f(3) = (3)^{3}  27(3) + 2 = 27  81 + 2 = 52
f(0) = (0)^{3}  27(0) + 2 = 2
f(4) = (4)^{3}  27(4) + 2 = 42
So the minimum value of f(x) on [0, 4] is 52 and its maximum value on [0, 4] is 56.
Important Notes on Extreme Value Theorem
 The extreme value theorem can also be stated as 'If a realvalued function f is continuous on [a, b], then f attains its maximum and minimum of [a, b].
 We can find the maximum and minimum values of a function by finding the critical points of the function using its derivative.
 The extreme value theorem can be proved using the contradiction and boundedness theorem.
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Extreme Value Theorem Examples

Example 1: Find the maximum and minimum values of f(x) = x^{4}  3x^{3}  1 on [2, 2].
Solution: Since f(x) is differentiable, so it is continuous on [2, 2]. So, we can apply extreme value theorem. Now, differentiate f(x).
f'(x) = 4x^{3}  9x^{2}
= x^{2}(4x  9)
Setting f'(x) = 0, we have
x^{2}(4x  9) = 0
⇒ x = 0, x = 9/4
So, 0 and 9/4 are the critical points. Since 9/4 does lie in the interval [2, 2], therefore we will only consider one critical point x = 0. Now, we will find the value of f(x) at x = 0, 2 and 2.
f(0) = 0^{4}  3(0)^{3}  1 = 1
f(2) = (2)^{4}  3(2)^{3}  1 = 16 + 24  1 = 39
f(2) = (2)^{4}  3(2)^{3}  1 = 16  24  1 = 9
So, the maximum value of f(x) = x^{4}  3x^{3}  1 on [2, 2] is 39 at x = 2 and minimum value is 9 at x = 2.
Answer: Maximum value = 39; Minimum value = 9

Example 2: Determine the maximum and minimum values of f(x) = sin x + cos x on [0, 2π] using the extreme value theorem.
Solution: f(x) = sin x + cos x on [0, 2π] is continuous. So we can apply extreme value theorem and find the derivative of f(x).
f'(x) = cos x  sin x
Setting f'(x) = 0, we have
cos x  sin x = 0
⇒ cos x = sin x
⇒ x = π/4, 5π/4 which lie in [0, 2π]
So, we will find the value of f(x) at x = π/4, 5π/4, 0 and 2π.
f(π/4) = sin (π/4) + cos (π/4) = 1/√2 + 1/√2 = √2
f(5π/4) = sin (5π/4) + cos (5π/4) = 1/√2  1/√2 = √2
f(0) = sin 0 + cos 0 = 0 + 1 = 1
f(2π) = sin 2π + cos 2π = 0 + 1 = 1
So, the maximum value of f(x) = sin x + cos x on [0, 2π] is √2 at x = π/4 and minimum value is √2 at x = 5π/4.
Answer: Maximum value = √2; Minimum value = √2
FAQs on Extreme Value Theorem
What is Extreme Value Theorem in Math?
The extreme value theorem is an important theorem in calculus that is used to find the maximum and minimum values of a continuous realvalued function in a closed interval.
What is Extreme Value Theorem Formula?
Mathematically, we can write the formula for the extreme value theorem as, f(c) ≤ f(x) ≤ f(d), ∀ x ∈ [a, b], where f is a continuous function on closed interval [a, b] and c, d lie in [a, b].
What is the Extreme Value Theorem Statement?
The extreme value theorem states that 'If a realvalued function f is continuous on a closed interval [a, b] (with a < b), then there exist two real numbers c and d in [a, b] such that f(c) is the minimum and f(d) is the maximum value of f(x).
How to Use the Extreme Value Theorem?
The extreme value theorem is used in proving the existence of the maximum and minimum values of a realvalued continuous function over a closed interval. Once the existence of maximum and minimum values is proved, we might be asked to determine those values using the derivative of the function and finding the critical points. We find the value of the function at critical points and the endpoints of the interval to the maximum and minimum values.
When Does Extreme Value Theorem Not Apply?
The extreme value theorem cannot be applied if the function is not continuous on the closed and bounded interval [a, b].
What is the Condition of Extreme Value Theorem?
The necessary condition of the extreme value theorem is that the function should be continuous on the closed and bounded interval [a, b].
How to Prove Extreme Value Theorem?
The extreme value theorem can be proved using the contradiction and boundedness theorem. We can prove the existence of the maximum value and similarly for the minimum value of the function.
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