Single-Variable-Calculus · Unit 11 · Video 4 · Interactive Practice

The Mean Value Theorem: Statement and Geometric Proof

IKey Formulas

Formula	Name	What you need
$\dfrac{f(b) - f(a)}{b - a} = f'(c)$ for some $c$ , $a < c < b$	Mean Value Theorem	$f$ continuous on $a \le x \le b$ and differentiable on $a < x < b$
$\dfrac{f(b) - f(a)}{b - a} = \dfrac{\Delta f}{\Delta x}$	Secant slope (average rate)	The two endpoint values
$f'(c)$	Tangent slope (instantaneous rate)	The derivative at one point
$\dfrac{3000 - 0}{6 - 0} = 500$	The trip: average speed in mph	$f(0) = 0$ , $f(6) = 3000$

Key Insight: Equal slopes mean parallel lines — somewhere strictly between $a$ and $b$ , a tangent line is parallel to the secant line.

IIWhere Is $c$ ?

For any interval $[a, b]$ , every point where the tangent is parallel to the secant is a valid $c$ — often more than one.

IIIThe Sliding-Line Proof

A line parallel to the secant, slid toward the graph from the side the graph bulges toward, first touches it strictly between $a$ and $b$ , along a tangent.

Graph Slide Position 0%

💡 This picture is a visual argument: proving that a first point of contact exists, and that the line there is a tangent, takes more analysis than this course covers.

IVWhere Differentiability Is Used

For $f(x) = |x|$ , an interval containing the corner has a secant slope that no derivative value matches.

💡 A corner exactly at an endpoint is allowed: the theorem only asks for a derivative strictly between $a$ and $b$ .

VQuiz Questions

Problem 1 · Find the Point c

Given: $f(x) = x^2$ on $1 \le x \le 3$ — find the value $c$ promised by the Mean Value Theorem.

✅ Correct! The secant slope is

4

and

f'(2) = 4

, with

1 < 2 < 3

❌ Close, but…

4

is the secant slope, the value

f'(c)

must equal. Now solve

2c = 4

for

c

❌ Not quite.

\sqrt{5}

solves

f(c) = 5

, the average of the endpoint heights. The theorem matches the slope

f'(c)

, not the height

f(c)

❌ Not quite. Compute the secant slope

\frac{f(3) - f(1)}{3 - 1}

, then solve

f'(c) = 2c

equal to it.

Show solution

$f$ is a polynomial, so it is continuous on $[1, 3]$ and differentiable on $(1, 3)$ . The secant slope is

\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4.

Since $f'(x) = 2x$ , set $2c = 4$ , so $c = 2$ , which lies strictly between $1$ and $3$ .

Problem 2 · Check the Hypotheses

Given: four functions on four intervals — which one is the Mean Value Theorem guaranteed to apply to?

✅ Correct! A polynomial is continuous and differentiable everywhere, so both hypotheses hold on any interval.

❌ Not quite.

|x|

is continuous, but

f'(0)

does not exist and

0

lies strictly between

-1

and

2

. The secant slope is

\tfrac{1}{3}

, while

f'

is only ever

-1

+1

❌ Not quite.

\frac{1}{x}

is not even defined at

x = 0

, which lies inside the interval, so continuity fails there.

❌ Not quite. Differentiability inside is fine, but

f

jumps at the endpoint

x = 1

: the curve heads toward

1

while

f(1) = 5

, so continuity on the closed interval fails.

Show solution

Check both hypotheses — continuity on $a \le x \le b$ and differentiability on $a < x < b$ — for each choice.

$|x|$ on $[-1, 2]$ : continuous, but not differentiable at the interior point $0$ .
$\frac{1}{x}$ on $[-1, 1]$ : undefined at $0$ , so not continuous.
The jump function: $\lim_{x \to 1^-} f(x) = 1 \ne 5 = f(1)$ , so not continuous at the endpoint.
$x^3 - 2x$ on $[-1, 2]$ : a polynomial, continuous and differentiable everywhere. ✓

Indeed $\frac{f(2) - f(-1)}{2 - (-1)} = \frac{4 - 1}{3} = 1$ and $f'(c) = 3c^2 - 2 = 1$ gives $c = 1$ , inside $(-1, 2)$ .

Problem 3 · Secant Slope, Then c

Given: $f(x) = x^3 - 3x$ on $0 \le x \le 3$ — find the secant slope and every $c$ the theorem allows.

What is the secant slope?

Which $c$ satisfy the conclusion?

✅ Correct!

f'(\sqrt{3}) = 3 \cdot 3 - 3 = 6

, and

0 < \sqrt{3} \approx 1.73 < 3

❌ Close, but…

18

is the rise

f(3) - f(0)

. Divide by the run

3 - 0

❌ Check the secant slope.

f(3) = 27 - 9 = 18

and

f(0) = 0

❌ Close, but…

-\sqrt{3}

solves the equation but lies outside

0 < c < 3

, so it does not count.

❌ Not quite.

c = 1

is where

f'(c) = 0

— a horizontal tangent. The tangent must match the secant slope

6

, not zero.

❌ Check your $c$ . Solve

f'(c) = 3c^2 - 3 = 6

and keep only solutions strictly between

0

and

3

Show solution

Step 1: Secant slope. $f(3) = 27 - 9 = 18$ , $f(0) = 0$ :

\frac{f(3) - f(0)}{3 - 0} = \frac{18}{3} = 6.

Step 2: Match the derivative. $f'(x) = 3x^2 - 3$ , so

3c^2 - 3 = 6 \;\Longrightarrow\; c^2 = 3 \;\Longrightarrow\; c = \pm\sqrt{3}.

Step 3: Keep the interior solutions. Only $c = \sqrt{3} \approx 1.732$ lies in $0 < c < 3$ .

Problem 4 · Two Speed Cameras

Given: a car passes two cameras $3$ miles apart, $2$ minutes apart — what speed does the Mean Value Theorem guarantee the car had at some instant in between?

✅ Correct! The average speed is

90

mph, so at some instant the speedometer read exactly

90

❌ Close, but…

3 / 2 = 1.5

is in miles per minute. Convert:

1.5 \times 60 = 90

mph.

❌ Not quite. That is exactly what the theorem rules out: position is differentiable in time, so the average rate is attained as an instantaneous rate.

❌ Not quite. The theorem promises some instant — at least one — not every instant. The car may speed up and slow down.

Show solution

Let $f(t)$ be the distance travelled after $t$ hours. Two minutes is $\tfrac{2}{60} = \tfrac{1}{30}$ hour, so the average speed is

\frac{f(t_2) - f(t_1)}{t_2 - t_1} = \frac{3}{1/30} = 90 \text{ mph}.

Position is continuous and differentiable in time, so by the Mean Value Theorem $f'(c) = 90$ for some instant $c$ between the cameras — whatever the speed limit says.

Solved: 0 / 4

The Mean Value Theorem: Statement and Geometric Proof

IKey Formulas

IIWhere Is ccc?

IIIThe Sliding-Line Proof

IVWhere Differentiability Is Used

VQuiz Questions

IIWhere Is $c$ ?