The Mean Value Theorem
State and prove the Mean Value Theorem and its special case, Rolle's Theorem. Understand the MVT as guaranteeing an instantaneous rate equal to the average rate, and apply it to justify estimates and prove inequalities.
A lesson of Calculus, in Applications of Derivatives, about 10 minutes.
By the end of this lesson, you will be able to state the Mean Value Theorem and Rolle's Theorem, apply them to existence proofs and function analysis, and understand their geometric meaning.
Rolle's Theorem
If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$.
Geometric meaning: if a smooth curve starts and ends at the same height, it must have at least one horizontal tangent somewhere between.
The Mean Value Theorem (MVT)
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c \in (a,b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$
Geometric meaning: there is at least one point where the tangent line is parallel to the secant line through $(a, f(a))$ and $(b, f(b))$.
Rolle's Theorem is the special case $f(a) = f(b)$ (secant slope = 0, so tangent slope = 0 somewhere).
Example 1 — Verifying MVT
$f(x) = x^3 - x$ on $[0, 2]$. Find $c$.
$\dfrac{f(2)-f(0)}{2-0} = \dfrac{6-0}{2} = 3$
Find $c$: $f'(c) = 3x^2 - 1 = 3 \implies x^2 = \frac{4}{3} \implies x = \frac{2}{\sqrt{3}} \approx 1.15$
This is in $(0,2)$ ✓. The tangent at $c = 2/\sqrt{3}$ is parallel to the secant.
Important Corollaries
Corollary 1: If $f'(x) = 0$ for all $x$ in $(a,b)$, then $f$ is constant on $(a,b)$.
Corollary 2: If $f'(x) = g'(x)$ for all $x$ in $(a,b)$, then $f(x) = g(x) + C$ for some constant $C$.
Corollary 3: If $f'(x) > 0$ on $(a,b)$, then $f$ is increasing on $(a,b)$.
Example 2 — Rolle's Theorem Application
Show $f(x) = x^4 - 2x^2 + 1$ has a zero of its derivative in $(-1, 1)$.
$f(-1) = 1 - 2 + 1 = 0 = f(1)$; $f$ is continuous and differentiable everywhere.
By Rolle's Theorem: $\exists c \in (-1,1)$ with $f'(c) = 0$. ✓
To locate that point explicitly: $f'(x) = 4x^3 - 4x = 4x(x^2-1) = 0$ at $x = 0, \pm 1$, so $c = 0$ is such a point inside $(-1, 1)$.
Example 3 — Proving Uniqueness
Prove $f(x) = x^3 + 5x + 2$ has exactly one real zero.
$f(0) = 2 > 0$ and $f(-1) = -1 - 5 + 2 = -4 < 0$, so a zero exists in $(-1, 0)$ by IVT.
Uniqueness: $f'(x) = 3x^2 + 5 > 0$ always. A function that is always increasing can have at most one zero. If there were two zeros $a < b$ with $f(a) = f(b) = 0$, Rolle's Theorem would give a $c$ with $f'(c) = 0$ — contradiction.