AP Calculus AB · Study guide

Using the Mean Value Theorem

AP Calculus · Analytical Applications of Differentiation

Using the Mean Value Theorem

Somewhere inside the interval, the instantaneous rate equals the average rate — provided both hypotheses hold.

key idea$\exists\,c\in(a,b):\ f'(c)=\frac{f(b)-f(a)}{b-a}$

The theorem

The Mean Value Theorem

Hypotheses — all must hold

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

What you may then conclude

There exists at least one $c$ in $(a,b)$ where the instantaneous rate equals the average rate: $f'(c)=\dfrac{f(b)-f(a)}{b-a}$. Geometrically, some tangent is parallel to the secant joining the endpoints.

Drop differentiability and the conclusion dies

$f(x)=\lvert x-2\rvert+1$ on $[0,4]$ is continuous, and $f(0)=f(4)=3$, so the secant is horizontal with slope $0$. But $f'$ is either $-1$ or $+1$ everywhere it exists — never $0$. The corner at $x=2$ voids the theorem.

MVT: some tangent matches the secant

corner → MVT does not apply

Left: a differentiable curve, where some interior tangent is parallel to the endpoint secant. Right: the same endpoint heights with a corner — no tangent ever matches the secant's slope.

Checking the hypotheses

As with every theorem question, the justification carries the marks. State both hypotheses, say why each holds, then invoke the theorem by name.

The three-sentence justification

Continuity on $[a,b]$ — and say why: “$f$ is a polynomial”, or check the domain for rational and root functions.
Differentiability on $(a,b)$ — note that this is the open interval, so endpoint behaviour never matters.
Invoke the theorem — “by the Mean Value Theorem there exists $c$ in $(a,b)$ with $f'(c)=\dots$”, giving the computed average rate.

Question type	What you do	Watch for
“Find all $c$”	compute the average rate, set $f'(c)$ equal, solve	discard roots outside $(a,b)$
“Justify that some $c$ exists”	verify both hypotheses and cite the theorem	no computation needed
“Explain why MVT does not apply”	name the failing hypothesis and where	a corner, cusp or break
“A car travelled…”	average speed over the trip	MVT guarantees that speed occurred

Row 2 asks for no arithmetic at all — only the hypothesis check. Computing $c$ there wastes time and earns nothing extra.

Worked examples

EX 1

Find every $c$

$f(x)=x^{2}$ on $[1,3]$

Verify the MVT applies to $f(x)=x^{2}$ on $[1,3]$ and find all values of $c$.

Check the hypotheses

$f$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$. Both hold.

Compute the average rate

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

Set the derivative equal

$f'(x)=2x$, so solve $2c=4$.

Solve

$c=2$.

Confirm it is interior

$2$ lies in the open interval $(1,3)$. ✓ So $c=2$ is the value the theorem promised.

EX 2

Justify without computing

existence only

$g$ is differentiable on $[2,8]$ with $g(2)=5$ and $g(8)=23$. Show some $c$ has $g'(c)=3$.

Check the hypotheses

Differentiable on $[2,8]$ implies continuous there, so both MVT hypotheses hold.

Compute the average rate

$\dfrac{g(8)-g(2)}{8-2}=\dfrac{23-5}{6}=3$.

Invoke the theorem

By the MVT there exists $c$ in $(2,8)$ with $g'(c)=3$.

Stop there

No formula for $g$ exists, and none is needed. The theorem guarantees existence, which is exactly what was asked.

Note the phrasing

“At least one” — there may be many such $c$, and the theorem never says how many.

EX 3

Why a hypothesis matters

the corner

Does the MVT apply to $f(x)=\lvert x-2\rvert+1$ on $[0,4]$?

Check continuity

Absolute value is continuous everywhere, so $f$ is continuous on $[0,4]$. ✓

Check differentiability

$f'$ does not exist at $x=2$, which lies inside $(0,4)$. ✗

Conclude about the theorem

The second hypothesis fails, so the MVT does not apply.

Show the conclusion genuinely fails

$f(0)=f(4)=3$, so the average rate is $0$. But $f'$ is $-1$ or $+1$ wherever it exists — never $0$.

The point

This is not a technicality. A single non-differentiable interior point destroys the guarantee entirely.

EX 4

The speeding-ticket argument

applied

A car passes two toll booths $180$ km apart in $90$ minutes. The limit is $100$ km/h. Prove it exceeded the limit.

Model it

Let $s(t)$ be distance travelled. It is continuous and differentiable — position cannot jump, and the car has a speedometer reading at every instant.

Compute the average speed

$\dfrac{180\text{ km}}{1.5\text{ h}}=120$ km/h.

Apply the MVT

There exists an instant $c$ in the interval with $s'(c)=120$ km/h.

Draw the conclusion

At that instant the car was travelling at $120$ km/h, exceeding the $100$ limit.

Note what makes this rigorous

The MVT converts an average over a trip into a guaranteed instantaneous fact. Without it, averaging $120$ would not prove any particular moment exceeded the limit.

When it fails

Closed for continuity, open for differentiability

$[a,b]$ and $(a,b)$ respectively. A function may fail to be differentiable at an endpoint and the theorem still applies.

A single interior corner voids it

Example 3 satisfies continuity perfectly and still fails, because differentiability breaks at one interior point.

Existence, never uniqueness or location

The theorem says at least one $c$ exists somewhere inside. It never says how many, or where.

Discard solutions outside the interval

Solving $f'(c)=m$ may give roots outside $(a,b)$. Only interior solutions count as the $c$ the theorem promises.

Course map. Topic 5.1 · LO FUN-1.B — justify conclusions about functions over an interval using the Mean Value Theorem (EK FUN-1.B.1).