Topic 5.1 Notes – Using the Mean Value Theorem

The Mean Value Theorem

Here’s the formal statement:

If a function $f$ is

• continuous on the closed interval $[a,b]$, and
• differentiable on the open interval $(a,b)$,

then there exists at least one number $c$ in $(a,b)$ such that

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

That fraction is the average rate of change on $[a,b]$.
The derivative $f'(c)$ is the instantaneous rate of change at one point.

So the theorem says: at some point inside the interval, the tangent line slope equals the secant line slope.

Here’s what that looks like visually:

Mean Value Theorem: secant and tangent lines

The dashed blue line represents the secant line between $(a, f(a))$ and $(b, f(b))$, and the magenta line is the tangent at $c$ with the same slope.

Two key ideas:

• It guarantees existence, not location (unless you solve for it).
• There could be more than one such $c$.

The Two Conditions You Must Check

Every justification problem lives or dies on these hypotheses. If they aren’t satisfied, you cannot use MVT.

1. Continuity on $[a,b]$

You need:

• No holes
• No jumps
• No vertical asymptotes
• The function defined at $a$ and $b$

Polynomials, exponentials, sine, cosine → continuous everywhere.
Piecewise functions → check endpoints carefully.

2. Differentiability on $(a,b)$

Inside the interval, there can’t be:

• Corners
• Cusps
• Vertical tangents
• Discontinuities

Remember: differentiability implies continuity. But continuity alone does not guarantee differentiability.

On a free-response question, you must state both conditions before concluding anything.

Using MVT to Justify a Derivative Value Exists

This is a very common quiz and FRQ setup.

You’ll be asked something like:
“Does there exist a value $c$ in $(1,5)$ such that $f'(c) = 3$?”

Here’s the structure that earns full credit:

1. State conditions clearly
   “Since $f$ is continuous on $[1,5]$ and differentiable on $(1,5)$, the Mean Value Theorem applies.”

2. Compute average rate of change

   $$\frac{f(5) - f(1)}{5 - 1}$$

3. Compare it to the claimed derivative value

   • If the result equals 3 → MVT guarantees such a $c$.
   • If it doesn’t → MVT cannot justify it.

Important detail students miss: you are comparing the average slope to the specific derivative value they gave you.

If they match exactly, you’re done.

Finding the Actual Value of $c$

Sometimes they want the number.

Example:
Let $f(x) = x^2 - 4x$ on $[0,6]$.

First, it’s a polynomial → continuous and differentiable everywhere.

Average rate of change:

$$\frac{f(6) - f(0)}{6 - 0} = \frac{(36 - 24) - 0}{6} = \frac{12}{6} = 2$$

Now compute derivative:

$$f'(x) = 2x - 4$$

Set it equal to 2:

$$2x - 4 = 2$$

$$2x = 6$$

$$x = 3$$

Since $3 \in (0,6)$, that value works.

If you get multiple solutions, only keep the ones inside the open interval.

Rolle’s Theorem as a Special Case

If $f(a) = f(b)$, then the average rate of change is 0.

So MVT guarantees some $c$ where:

$$f'(c) = 0$$

That’s called Rolle’s Theorem.

Geometrically, it means if the function starts and ends at the same height, it must have a horizontal tangent somewhere in between.

What MVT Tells You Conceptually

It lets you conclude behavior without knowing the exact point.

• If overall change is positive → derivative is positive somewhere.
• If overall change is negative → derivative is negative somewhere.
• If endpoints match → derivative is zero somewhere.

That’s powerful. You don’t need the full graph. Just the conditions and endpoint values.