The Mean Value Theorem
- Calculus 3.5
Rolle’s Theorem
Rolle’s Theorem is a Lemma that serves as a foundation for the Mean Value Theorem.
Let $f$ be a continuous function on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. If $f(a) = f(b)$, then there exists at least one point $c \in (a, b)$ such that $f^\prime(c) = 0$.
Proof
Being continuous, $f$ has absolute maximum and minimum values on $[a, b]$ by the Extreme Value Theorem. These can occur only at interior points where $f^\prime$ is zero, or at the endpoints $a$ and $b$. If either the maximum or minimum occurs at a point $c\in(a,b)$, then we’ve found the desired point where $f^\prime(c) = 0$. If both the maximum and minimum occur at the endpoints, then since $f(a) = f(b)$, $f$ becomes constant on $[a, b]$. Therefore, $f^\prime(c) = 0$ for all $c \in (a, b)$.
Mean Value Theorem
Mean Value Theorem(MVT) states that there exists a point where the tangent line is parallel to the secant line that joins the endpoints of the interval.
Let $f$ be a continuous function on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. Then there exists at least one point $c \in (a, b)$ such that:
\[ \frac{f(b) - f(a)}{b - a} = f^\prime(c) \]
Proof
Define the function $g$ as:
\[ g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a} (x - a) \]
Then $g$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Also, $g(a) = g(b) = 0$. By Rolle’s Theorem, there exists a point $c \in (a, b)$ such that $g^\prime(c) = 0$. Calculating $g^\prime(x)$ gives:
\[ g^\prime(x) = f^\prime(x) - \frac{f(b) - f(a)}{b - a} \]
Setting $g^\prime(c) = 0$ leads to:
\[ \frac{f(b) - f(a)}{b - a} = f^\prime(c) \]
Corollary
- If $f^\prime(x)=0$ for all $x \in (a, b)$, then $f$ is constant on $(a, b)$.
- If $f^\prime(x)=g^\prime(x)$ for all $x \in (a, b)$, then $f-g$ is constant on $(a, b)$.
