Mean Value Theorem and Rolle's Theorem

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $c \in (a,b)$ such that:

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

Interpretation: At some point, the instantaneous rate equals the average rate.

Special case: if $f(a) = f(b)$, then there exists $c$ where $f'(c) = 0$.

$f(x) = x^3 - 3x$ on $[0, 2]$.

Average rate: $\frac{f(2)-f(0)}{2} = \frac{2-0}{2} = 1$.

$f'(x) = 3x^2 - 3 = 1$ → $x^2 = \frac{4}{3}$ → $x = \frac{2}{\sqrt{3}} \approx 1.15$ ✓ (in $(0,2)$).