Course Content
AP Calculus BC

Mean Value Theorem (For Derivatives)


What is the Mean Value Theorem?

The Mean Value Theorem (MVT) connects the average rate of change of a function to its instantaneous rate of change at some point within an interval.

In simple terms:
Somewhere between two points on a smooth curve, there is at least one point where the tangent line is parallel to the secant line joining those two points.


The Formal Statement

Let $$f(x)$$ be a function that satisfies two conditions:

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

  2. $$f(x)$$ is differentiable on the open interval (a,b).

Then, there exists at least one number c∈(a,b) such that:


Understanding the Formula

  • The right-hand side, $$\frac{f(b) – f(a)}{b – a}$$, represents the average rate of change of $$f(x)$$ over [a,b].

  • The left-hand side, $$f^\prime(x)$$, represents the instantaneous rate of change (slope of the tangent) at x=c.

Thus, MVT says:

There exists a point cc where the instantaneous rate of change equals the average rate of change.


Visual Interpretation

Imagine a smooth curve from point A(a , f(a)) to .

  • The secant line connects A and B.

  • The tangent line at some point C(c , f(c)) is parallel to that secant line.

That’s the Mean Value Theorem in action!


Example 1

Find the point cc that satisfies the Mean Value Theorem for:

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

Step 1. Check the conditions:

$$f(x) = x^{2}$$ is continuous and differentiable everywhere. 

Step 2. Compute the average rate of change:

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

Step 3. Find c such that $$f^\prime(c) = 4$$.

$$f^\prime(x) = 2x$$ then $$2c = 4 \longrightarrow c=2$$

at c=2, the tangent line is parallel to the secant line between and x=3.


Example 2

Let $$f(x) = \sin(x)$$ on $$[0,\pi]$$.

$$\frac{f(\pi) – f(0)}{\pi – 0} = \frac{0}{\pi} = 0$$

Find c such that $$f^\prime(c)=0$$.

$$c = \frac{\pi}{2}$$satisfies the Mean Value Theorem.