Mean Value Theorem and Extreme Value Theorem
State and apply the EVT (a continuous function on a closed interval attains its absolute extrema) and MVT. Use the MVT to justify the existence of critical points and to estimate function values on AP free-response problems.
A lesson of AP Calculus AB, in Analytical Applications of Differentiation, about 10 minutes.
By the end of this lesson, you will be able to state and apply the Mean Value Theorem and Extreme Value Theorem, and explain their significance for understanding function behavior.
The Extreme Value Theorem (EVT)
Statement: If $f$ is continuous on the closed interval $[a, b]$, then $f$ attains both a maximum value and a minimum value on $[a, b]$.
Why "closed"? On an open interval, the function could approach but never reach an extreme value. Example: $f(x) = 1/x$ on $(0, 1)$ has no maximum.
Why "continuous"? A discontinuous function could have a "hole" at the potential maximum.
The EVT guarantees existence — it tells us extrema exist but doesn't tell us where. We use derivatives to find them.
Fermat's Theorem
If $f$ has a local maximum or minimum at $x = c$ (an interior point of the domain), and $f'(c)$ exists, then $f'(c) = 0$.
Corollary: The only candidates for local extrema are:
- Points where $f'(c) = 0$ (critical points)
- Points where $f'(c)$ doesn't exist (corner, cusp, vertical tangent)
The Mean Value Theorem (MVT)
Statement: If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $c \in (a, b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$
Interpretation: There is a point $c$ in $(a, b)$ where the instantaneous rate of change (slope of the tangent) equals the average rate of change (slope of the secant line) over $[a, b]$.
Geometric meaning: The tangent at $c$ is parallel to the secant line from $(a, f(a))$ to $(b, f(b))$.
Applying the MVT
Example: Verify the MVT applies to $f(x) = x^2$ on $[1, 3]$ and find $c$.
$f$ is a polynomial → continuous and differentiable on all of $\mathbb{R}$ ✓
$\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4$
$f'(x) = 2x$; set $2c = 4$ → $c = 2$ ∈ $(1, 3)$ ✓
Consequences of the MVT
Constant Function: If $f'(x) = 0$ for all x in $(a, b)$, then $f$ is constant on $[a, b]$.
Increasing/Decreasing: If $f'(x) > 0$ on $(a, b)$, then $f$ is increasing on $[a, b]$. If $f'(x) < 0$, decreasing.