# Mean Value Theorem (MVT): Statement, Graph and Examples

> The mean value theorem: a smooth function on [a, b] has a point where the tangent slope equals the average slope. Statement, graph, counterexamples, examples.

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

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

## At a glance

- **Hypotheses:** $f$ continuous on $[a, b]$ and differentiable on $(a, b)$
- **Conclusion:** Some $c$ in $(a, b)$ has $f'(c) = \frac{f(b) - f(a)}{b - a}$
- **Geometric meaning:** Some tangent line is parallel to the secant line through the endpoints
- **Physical meaning:** At some instant, instantaneous velocity equals average velocity
- **Special case:** Rolle's theorem, when $f(a) = f(b)$
- **Course:** Calculus 1

## Key takeaways

- Somewhere between $a$ and $b$, the instantaneous rate of change equals the average rate of change.
- The function must be continuous on $[a, b]$ and differentiable on $(a, b)$. A corner or a jump can break the conclusion.
- The MVT turns bounds on $f'$ into bounds on how much $f$ can change.
- If $f'(x) = 0$ on an interval, the MVT proves $f$ is constant there.

The **mean value theorem** (MVT) says that for a smooth function, the average rate of change over an interval is achieved as an instantaneous rate of change at some point inside it. Geometrically, some tangent line is parallel to the secant line joining the endpoints.

## Statement of the mean value theorem

Let $f$ satisfy both hypotheses:

1. $f$ is **continuous** on the closed interval $[a, b]$;
2. $f$ is **differentiable** on the open interval $(a, b)$.

Then there is at least one number $c$ in $(a, b)$ such that

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Equivalently, $f(b) - f(a) = f'(c)(b - a)$. This second form is the one used in proofs: it converts information about the derivative into information about the function.

## What the MVT means

The fraction $\frac{f(b) - f(a)}{b - a}$ is the slope of the secant line, the **average** rate of change. $f'(c)$ is the slope of the tangent, an **instantaneous** rate of change. The MVT says the two agree somewhere. If your average speed on a trip was 60 mph, at some instant you were going exactly 60 mph.

```graph
title: 'Mean value theorem: tangent to y = x³ parallel to the secant on [0, 2]'
caption: 'The secant from $(0, 0)$ to $(2, 8)$ has slope $4$. At $c = \frac{2}{\sqrt{3}} \approx 1.155$ the tangent line has the same slope.'
x: [-0.5, 2.5]
y: [-4, 10]
functions:
  - expr: x^3
    label: y = x³
  - expr: 4x
    label: secant, slope 4
    domain: [0, 2]
  - expr: 4x - 3.0792
    label: tangent at c
    dashed: true
points:
  - [0, 0, 'a = 0']
  - [2, 8, 'b = 2']
  - [1.1547, 1.5396, 'c ≈ 1.155']
```

## Why each hypothesis matters

**Differentiability.** Take $f(x) = |x|$ on $[-1, 2]$. It is continuous, and the secant slope is $\frac{2 - 1}{2 - (-1)} = \frac{1}{3}$. But $f'(x)$ is $-1$ for $x < 0$ and $1$ for $x > 0$, never $\frac{1}{3}$. The corner at $0$ is the culprit.

**Continuity.** Take $f(x) = \frac{1}{x}$ on $[-1, 1]$. The secant slope is $\frac{1 - (-1)}{2} = 1$, yet $f'(x) = -\frac{1}{x^2}$ is always negative. The function is not continuous at $0$, so the MVT does not apply.

## Worked examples

**Example 1: find $c$ for $f(x) = x^3$ on $[0, 2]$.**

**Step 1.** $f$ is a polynomial, so both hypotheses hold.

**Step 2.** Average slope: $\frac{f(2) - f(0)}{2 - 0} = \frac{8}{2} = 4$.

**Step 3.** Solve $f'(c) = 3c^2 = 4$, giving $c = \pm\frac{2}{\sqrt{3}}$. Only $c = \frac{2}{\sqrt{3}} \approx 1.155$ lies in $(0, 2)$.

**Example 2: find $c$ for $f(x) = \sqrt{x}$ on $[1, 4]$.**

$f$ is continuous on $[1, 4]$ and differentiable on $(1, 4)$. The average slope is $\frac{2 - 1}{4 - 1} = \frac{1}{3}$. Solve

$$\frac{1}{2\sqrt{c}} = \frac{1}{3} \implies \sqrt{c} = \frac{3}{2} \implies c = \frac{9}{4}$$

and $\frac{9}{4} = 2.25$ is in $(1, 4)$.

**Example 3: bounding a function.** Suppose $f(0) = 1$ and $f'(x) \le 3$ for all $x$. How large can $f(2)$ be?

By the MVT on $[0, 2]$, $f(2) - f(0) = f'(c) \cdot 2$ for some $c$. Since $f'(c) \le 3$,

$$f(2) \le 1 + 3 \cdot 2 = 7$$

You learn something about $f(2)$ without knowing the function. This is how the MVT is used in proofs and in [linear approximation](/blog/linear-approximation) error estimates.

## Key consequences

- **Zero derivative means constant.** If $f'(x) = 0$ on an interval, then for any two points $f(x_2) - f(x_1) = 0 \cdot (x_2 - x_1) = 0$, so $f$ is constant.
- **Same derivative means differ by a constant.** If $f' = g'$, then $f - g$ is constant. This is why every [antiderivative](/blog/antiderivatives) has a $+ C$.
- **Sign of $f'$ gives monotonicity.** If $f' > 0$ on an interval, $f$ is increasing there, the basis of the first derivative test.

## Common mistakes

1. **Answering with a $c$ outside $(a, b)$.** Solve $f'(c) = \text{slope}$, then discard roots not in the open interval.
2. **Skipping the hypothesis check.** For functions with roots, absolute values or fractions, check continuity and differentiability first.
3. **Thinking the MVT locates $c$.** It only guarantees one exists. In many applications, as in Example 3, you never find $c$ at all.

## Related theorems

The MVT is proved from [Rolle's theorem](/theorems/rolles-theorem) by subtracting the secant line from $f$. Its integral twin is the [mean value theorem for integrals](/theorems/mean-value-theorem-for-integrals), and [Taylor's theorem](/theorems/taylors-theorem) is a higher-order version. For more applications and practice, read the full [mean value theorem guide](/blog/mean-value-theorem).

## Frequently asked questions

### What does the mean value theorem say?

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there is a number $c$ in $(a, b)$ where $f'(c) = \frac{f(b) - f(a)}{b - a}$. The slope of the tangent at $c$ equals the slope of the secant line from $a$ to $b$.

### How do you find c in the mean value theorem?

Compute the average rate of change $\frac{f(b) - f(a)}{b - a}$, set $f'(c)$ equal to it, solve for $c$, and keep only the solutions that lie in the open interval $(a, b)$.

### What is a real-life example of the mean value theorem?

If you drive 120 miles in 2 hours, your average speed is 60 mph, so the MVT says your speedometer read exactly 60 mph at some moment during the trip.

### Why is differentiability required in the mean value theorem?

Without it, the graph can have a corner and no tangent line matches the secant slope. For $f(x) = |x|$ on $[-1, 2]$ the secant slope is $\frac{1}{3}$, but the derivative is only ever $-1$ or $1$.