# Extreme Value Theorem: Statement, Counterexamples, Examples

> The extreme value theorem: a continuous function on a closed interval [a, b] has an absolute max and min. Statement, graph, counterexamples and examples.

Source: https://practicecalculus.com/theorems/extreme-value-theorem
Updated: 2026-11-03

$$ f(c) \le f(x) \le f(d) \text{ for all } x \in [a, b] $$

## At a glance

- **Hypotheses:** $f$ is continuous on a closed, bounded interval $[a, b]$
- **Conclusion:** $f$ has an absolute minimum $f(c)$ and an absolute maximum $f(d)$ for some $c, d$ in $[a, b]$
- **Type:** Existence theorem: it guarantees extreme values, not their location
- **Where to look:** Critical points inside $(a, b)$ and the endpoints $a$ and $b$
- **Used for:** The closed interval method, optimization
- **Course:** Calculus 1

## Key takeaways

- A continuous function on a closed interval $[a, b]$ always reaches a highest and a lowest value.
- Both conditions matter: drop continuity or the closed interval and a max or min can fail to exist.
- The extremes occur at critical points or at the endpoints, which is why the closed interval method works.

The **extreme value theorem** (EVT) guarantees that a continuous function on a closed interval has a highest point and a lowest point. It is the reason optimization problems on a closed interval always have an answer, and it tells you exactly where to look for it.

## Statement of the extreme value theorem

If $f$ is **continuous** on the **closed, bounded interval** $[a, b]$, then there exist numbers $c$ and $d$ in $[a, b]$ such that

$$f(c) \le f(x) \le f(d) \quad \text{for all } x \in [a, b]$$

So $f(c)$ is the **absolute minimum** of $f$ on $[a, b]$ and $f(d)$ is the **absolute maximum**. The points $c$ and $d$ may be endpoints or interior points, and there may be more than one of each.

## What the EVT means

A continuous graph over a closed interval is an unbroken piece of curve with both ends included. Such a piece cannot run off to infinity, and it cannot creep toward a height without reaching it, so it has a top and a bottom.

```graph
title: 'Extreme value theorem: y = x³ − 3x on [0, 2.5] reaches its max and min'
caption: 'On $[0, 2.5]$ the minimum $-2$ occurs at the critical point $x = 1$ and the maximum $8.125$ occurs at the endpoint $x = 2.5$.'
x: [-0.5, 3]
y: [-3, 9]
functions:
  - expr: x^3 - 3x
    label: y = x³ − 3x
    domain: [0, 2.5]
points:
  - [1, -2, 'min (1, −2)']
  - [2.5, 8.125, 'max (2.5, 8.125)']
  - [0, 0, '(0, 0)']
vertical:
  - {x: 0, label: 'a = 0'}
  - {x: 2.5, label: 'b = 2.5'}
```

## Why each hypothesis matters

**The interval must be closed.** $f(x) = x$ on the open interval $(0, 1)$ is continuous, but it has no maximum: for any value you pick, a slightly larger $x$ gives a larger output. The would-be maximum $1$ lives at the excluded endpoint.

**The interval must be bounded.** $f(x) = x$ on $[0, \infty)$ has a minimum at $0$ but no maximum, because it grows forever.

**The function must be continuous.** Define $f(x) = x$ for $0 \le x < 1$ and $f(1) = 0$. On $[0, 1]$ the values get closer and closer to $1$ but never reach it, so there is no maximum. The jump at $x = 1$ is to blame. A vertical asymptote, as in $\frac{1}{x}$ on $[-1, 1]$ with $f(0)$ defined as $0$, also destroys the maximum and the minimum.

These conditions are sufficient, not necessary. Some discontinuous functions still have extreme values. The EVT just guarantees them when the hypotheses hold.

## The closed interval method

The EVT says the extremes exist. [Fermat's theorem](/theorems/fermats-theorem) says an interior extreme must be at a critical point. Together they give a complete recipe:

1. Check that $f$ is continuous on $[a, b]$.
2. Find the critical points of $f$ in $(a, b)$: where $f'(x) = 0$ or $f'(x)$ does not exist.
3. Evaluate $f$ at each critical point and at $a$ and $b$.
4. The largest value is the absolute maximum, the smallest the absolute minimum.

## Worked examples

**Example 1: $f(x) = x^3 - 3x$ on $[0, 2.5]$.**

**Step 1.** $f$ is a polynomial, so it is continuous and the EVT applies.

**Step 2.** $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$, which is zero at $x = \pm 1$. Only $x = 1$ lies in $(0, 2.5)$.

**Step 3.** Evaluate: $f(0) = 0$, $f(1) = 1 - 3 = -2$ and $f(2.5) = 15.625 - 7.5 = 8.125$.

**Step 4.** The absolute minimum is $-2$ at $x = 1$ and the absolute maximum is $8.125$ at $x = 2.5$. Notice the maximum is at an endpoint, where $f'$ is not zero. That is why you must always check the endpoints.

**Example 2: $f(x) = x - 2\sin x$ on $[0, \pi]$.**

$f'(x) = 1 - 2\cos x = 0$ gives $\cos x = \frac{1}{2}$, so $x = \frac{\pi}{3}$ in $(0, \pi)$. Then

$$f(0) = 0, \qquad f\left(\tfrac{\pi}{3}\right) = \tfrac{\pi}{3} - \sqrt{3} \approx -0.685, \qquad f(\pi) = \pi$$

The absolute minimum is $\frac{\pi}{3} - \sqrt{3}$ at $x = \frac{\pi}{3}$ and the absolute maximum is $\pi$ at $x = \pi$.

**Example 3: when the EVT does not apply.** Does $f(x) = \frac{1}{x}$ have a maximum on $(0, 2]$? The interval is not closed, so the EVT makes no promise. Indeed, $f(x) \to \infty$ as $x \to 0^+$, so there is no maximum. There is a minimum, $f(2) = \frac{1}{2}$, which shows the EVT's conclusion can still partly hold.

## Common mistakes

1. **Forgetting the endpoints.** Setting $f'(x) = 0$ only finds interior candidates. The answer is often at $a$ or $b$.
2. **Using critical points outside the interval.** In Example 1, $x = -1$ is a critical point but not in $[0, 2.5]$, so it is ignored.
3. **Applying the EVT on open or infinite intervals.** There, look at limits at the edges instead, as in [optimization problems](/blog/optimization-problems).

## Related theorems

The EVT is used to prove [Rolle's theorem](/theorems/rolles-theorem) and therefore the [mean value theorem](/theorems/mean-value-theorem). It shares its continuity hypothesis with the [intermediate value theorem](/theorems/intermediate-value-theorem). For sketching and classifying extrema, see [curve sketching with derivatives](/blog/curve-sketching-with-derivatives).

## Frequently asked questions

### What does the extreme value theorem say?

If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains an absolute maximum and an absolute minimum on that interval. There are points $c$ and $d$ in $[a, b]$ with $f(c) \le f(x) \le f(d)$ for every $x$ in the interval.

### Why does the extreme value theorem need a closed interval?

On an open interval the function can approach a value at the edge without ever reaching it. For example $f(x) = x$ on $(0, 1)$ gets arbitrarily close to $0$ and $1$ but has no maximum or minimum.

### How do you find the absolute max and min on a closed interval?

Find the critical points of $f$ in $(a, b)$, evaluate $f$ at those points and at both endpoints, and compare. The largest value is the absolute maximum and the smallest is the absolute minimum.

### Can the maximum and minimum occur at the same point?

Only if the function is constant on the interval. Then every point is both an absolute maximum and an absolute minimum.