Does the Extreme Value Theorem tell me how to find the max and min?

No, EVT only guarantees that a max and min exist if the function is continuous on a closed interval. To find them, use the Closed Interval Method: find critical points, evaluate the function at those points and the endpoints, then compare values.

What happens if the interval is open, like (a, b) instead of [a, b]?

EVT does not apply to open intervals. A continuous function on an open interval might approach extreme values without ever reaching them. For example, f(x) = x on (0, 1) has no minimum or maximum.

Can I use the Extreme Value Theorem if my function has a jump discontinuity?

No. EVT requires the function to be continuous on the entire closed interval [a, b]. If there is even one jump, hole, or vertical asymptote in that interval, EVT does not apply and the extrema may not exist.

Why do I need to check endpoints if I found all the critical points?

The extreme values can occur at critical points where f'(x) = 0, but they can also occur at the endpoints of the interval. You must evaluate the function at every critical point and both endpoints to guarantee you have found the true max and min.

Extreme Value Theorem

If a function is continuous on a closed interval, it must attain both its maximum and minimum values somewhere on that interval.

f \text{ continuous on } [a,b] \implies f \text{ attains a max and a min}

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What each symbol means

What Extreme Value Theorem takes

Symbol	Meaning
$f$	The function you're analyzing; it must be continuous on [a,b] for the theorem to apply, or it might miss a maximum or minimum value.
$a$	The left endpoint of the closed interval; a real number where the interval begins, included in the domain.
$b$	The right endpoint of the closed interval; a real number where it ends (must satisfy $a < b$ ), included in the domain.

When to use it

Reach for the Extreme Value Theorem when you need to guarantee that a continuous function has both max and min values on a closed interval.

Level

Usually taught in: Calculus I

Worked examples

1. Find extreme values of a quadratic on a closed interval

Problem

Find the extreme values of

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

[-1, 2]

$f(x) = 2x^2 - 3x + 1 \text{ is continuous on } [-1, 2]$
The function is a polynomial, so it is continuous everywhere, including on the closed interval $[-1, 2]$ . By the Extreme Value Theorem, $f$ must attain both a maximum and minimum on this interval.
$f'(x) = 4x - 3$
Differentiate $f$ to find critical points.
$4x - 3 = 0 \Rightarrow x = \frac{3}{4}$
Solve $f'(x) = 0$ to find the critical point $x = 3/4$ , which lies in $[-1, 2]$ .
$f(-1) = 2(-1)^2 - 3(-1) + 1 = 2 + 3 + 1 = 6$
Evaluate $f$ at the left endpoint $x = -1$ .
$f(\tfrac{3}{4}) = 2(\tfrac{3}{4})^2 - 3(\tfrac{3}{4}) + 1 = \tfrac{9}{8} - \tfrac{9}{4} + 1 = -\tfrac{1}{8}$
Evaluate $f$ at the critical point $x = 3/4$ .
$f(2) = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3$
Evaluate $f$ at the right endpoint $x = 2$ .

Answer: $\max f = 6 \text{ at } x = -1 \text{ and } \min f = -\tfrac{1}{8} \text{ at } x = \tfrac{3}{4}$

Since $f$ is continuous on the closed interval $[-1, 2]$ , the Extreme Value Theorem guarantees that extreme values exist. The Closed Interval Method finds them by comparing function values at all critical points and endpoints. The largest value is 6 and the smallest is $-1/8$ .

2. Find extrema with multiple critical points

Problem

Find the extreme values of

g(x) = x^3 - 3x

[-2, 1]

$g(x) = x^3 - 3x \text{ is continuous on } [-2, 1]$
The function is a polynomial, so it is continuous on the closed interval $[-2, 1]$ . By the Extreme Value Theorem, $g$ attains both a maximum and minimum.
$g'(x) = 3x^2 - 3$
Differentiate $g$ to find critical points.
$3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$
Solve $g'(x) = 0$ to find critical points. Both $x = -1$ and $x = 1$ lie in $[-2, 1]$ .
$g(-2) = (-2)^3 - 3(-2) = -8 + 6 = -2$
Evaluate $g$ at the left endpoint $x = -2$ .
$g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$
Evaluate $g$ at the critical point $x = -1$ .
$g(1) = (1)^3 - 3(1) = 1 - 3 = -2$
Evaluate $g$ at the critical point $x = 1$ .
$\text{Comparing: } g(-2) = -2, \, g(-1) = 2, \, g(1) = -2$
Identify the maximum and minimum by comparing the function values at all critical points and endpoints.

Answer: $\max g = 2 \text{ at } x = -1 \text{ and } \min g = -2 \text{ at } x = -2 \text{ and } x = 1$

This example illustrates that a continuous function on a closed interval not only has extreme values (guaranteed by EVT), but they can occur at critical points or at endpoints. Here, the maximum occurs at a critical point, while the minimum occurs at two different endpoints.

3. Find altitude extrema for a cyclist's hill training

Problem

A cyclist riding up and down a hill has altitude

a(t) = 100 - 50(t - 1)^2

meters after traveling

t

kilometers, for

0 \le t \le 2

. Find her minimum and maximum altitudes on this route.

$a(t) = 100 - 50(t-1)^2 \text{ is continuous on } [0, 2]$
The function $a(t)$ is a polynomial in $t$ , so it is continuous on the closed interval $[0, 2]$ . By the Extreme Value Theorem, the altitude attains both a maximum and minimum on this interval.
$a'(t) = -50 \cdot 2(t - 1) = -100(t - 1)$
Differentiate $a(t)$ using the chain rule to find critical points.
$-100(t - 1) = 0 \Rightarrow t = 1$
Solve $a'(t) = 0$ to find the critical point $t = 1$ km, which lies in $[0, 2]$ .
$a(0) = 100 - 50(0 - 1)^2 = 100 - 50 = 50$
Evaluate $a$ at the starting point $t = 0$ km.
$a(1) = 100 - 50(1 - 1)^2 = 100 - 0 = 100$
Evaluate $a$ at the critical point $t = 1$ km, which is the hilltop.
$a(2) = 100 - 50(2 - 1)^2 = 100 - 50 = 50$
Evaluate $a$ at the endpoint $t = 2$ km.

Answer: $\max a = 100 \text{ m at } t = 1 \text{ km and } \min a = 50 \text{ m at } t = 0 \text{ and } t = 2 \text{ km}$

The Extreme Value Theorem guarantees that this continuous function on a closed interval has extreme altitudes. At the hilltop ( $t = 1$ km), the cyclist reaches her maximum altitude of 100 meters. She returns to her minimum altitude of 50 meters at both the start and end of the route, illustrating that extrema can occur at multiple points.

Common mistakes

Where Extreme Value Theorem usually goes wrong

The mistake	Why it is wrong	The fix
Assuming EVT applies to an open interval like $(a, b)$	EVT requires a closed interval $[a, b]$ ; on an open interval, the function might approach but never attain extreme values.	Always check that the interval is closed $[a, b]$ with both endpoints included.
Assuming EVT applies even though the function has a discontinuity on $[a, b]$	EVT requires continuity; if there is a jump or hole in $[a, b]$ , the function might not attain its theoretical extremes.	Verify the function is continuous throughout the entire closed interval before invoking EVT.
Using EVT to find extreme values without checking critical points or endpoints	EVT guarantees extrema exist, but it does not tell you where they are; you still must evaluate at endpoints and critical points.	After confirming EVT applies, find critical points by solving $f'(x) = 0$ and evaluate $f$ at all critical points and both endpoints.

Tips and when to use something else

EVT only promises extrema exist; use the Closed Interval Method (evaluate $f$ at all critical points and endpoints) to find them.
Always verify both conditions: is the function continuous on the interval, and is the interval closed $[a, b]$ ?
If your function is discontinuous or your interval is open, do not use EVT—instead find the supremum and infimum (which may not be attained).
For functions on unbounded intervals, you cannot use EVT; instead use derivatives and limits to analyze behavior, or consider the Intermediate Value Theorem for different questions.

Related formulas

ContinuityContinuity tells you when a function has no jumps or breaks, so you can safely evaluate limits and apply theorems about function behavior.Intermediate Value TheoremIntermediate Value Theorem states that a continuous function on a closed interval must cross any value between its endpoints.Removable DiscontinuityA removable discontinuity is a hole in a graph where the limit exists but the function value differs; it can be fixed by redefining that point.

Frequently asked questions

Does the Extreme Value Theorem tell me how to find the max and min?: No, EVT only guarantees that a max and min exist if the function is continuous on a closed interval. To find them, use the Closed Interval Method: find critical points, evaluate the function at those points and the endpoints, then compare values.
What happens if the interval is open, like $(a, b)$ instead of $[a, b]$ ?: EVT does not apply to open intervals. A continuous function on an open interval might approach extreme values without ever reaching them. For example, $f(x) = x$ on $(0, 1)$ has no minimum or maximum.
Can I use the Extreme Value Theorem if my function has a jump discontinuity?: No. EVT requires the function to be continuous on the entire closed interval $[a, b]$ . If there is even one jump, hole, or vertical asymptote in that interval, EVT does not apply and the extrema may not exist.
Why do I need to check endpoints if I found all the critical points?: The extreme values can occur at critical points where $f'(x) = 0$ , but they can also occur at the endpoints of the interval. You must evaluate the function at every critical point and both endpoints to guarantee you have found the true max and min.

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Reviewed 2026-09-18