What the Mean Value Theorem Says and How to Apply It

The theorem matches an average slope to an instantaneous slope

If $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then at least one number $c\in(a,b)$ satisfies

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

The fraction on the right is the slope of the secant line through the two endpoints. The derivative on the left is the slope of a tangent line inside the interval. Geometrically, the theorem guarantees at least one tangent parallel to that secant.

Check the hypotheses before solving

The closed-interval continuity condition rules out jumps and holes between the endpoints. Differentiability on the open interval rules out corners, cusps, vertical tangents, and other points where the required finite derivative does not exist.

If either condition fails, the equation for $c$ may still happen to have a solution, but the Mean Value Theorem no longer guarantees one.

A standard solution process

1. Verify continuity on $[a,b]$ and differentiability on $(a,b)$.
2. Compute the average rate of change $[f(b)-f(a)]/(b-a)$.
3. Differentiate $f$.
4. Solve $f'(c)=[f(b)-f(a)]/(b-a)$.
5. Keep only solutions inside $(a,b)$.

Example with a different function

For $f(x)=x^2$ on $[1,3]$, the hypotheses hold because every polynomial is continuous and differentiable. The secant slope is

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

Since $f'(x)=2x$, solve $2c=4$ to get

$$\boxed{c=2}.$$

The theorem guarantees existence, not uniqueness. Some functions have more than one interior point with the required tangent slope.