The Mean Value Theorem

The Mean Value Theorem (MVT) says: if a function is differentiable on an interval, then somewhere in that interval its instantaneous rate of change equals its average rate of change.

Prerequisites

• Derivatives, Differentials, and the Chain Rule --- what $f^{\prime }(x)$ means
• Continuity on closed intervals, differentiability on open intervals

Notation

• $f$ --- a function continuous on $[a,b]$ and differentiable on $(a,b)$
• $f^{\prime }(c)$ --- the derivative (instantaneous rate of change) at point $c$
• $\frac{f(b)-f(a)}{b-a}$ --- the average rate of change (slope of the secant line from $a$ to $b$)

Statement

The dashed blue line is the secant (average slope over $[a,b]$). The red line is the tangent at $c$. MVT guarantees at least one $c$ where these are parallel --- but not where.

Hypothesis: $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

Thesis: There exists $c\in (a,b)$ such that

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

Equivalently, multiplying both sides by $(b-a)$:

$f(b)-f(a)=f^{\prime }(c)\cdot (b-a)$

Proof

The proof constructs a helper function that turns the problem into an application of Rolle’s theorem (a special case of MVT where $f(a)=f(b)$).

Step 1. Define $h(x)=f(x)-L(x)$, where $L(x)$ is the secant line from $(a,f(a))$ to $(b,f(b))$.

$L(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$

Step 2. Observe that $h(a)=0$ and $h(b)=0$ (both endpoints lie on the secant line, so $f-L=0$ there).

Step 3. $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$ (since $f$ is, and $L$ is a polynomial). Apply Rolle’s theorem: there exists $c\in (a,b)$ with $h^{\prime }(c)=0$.

Step 4. Compute $h^{\prime }(x)$.

$h^{\prime }(x)=f^{\prime }(x)-\frac{f(b)-f(a)}{b-a}$

Step 5. Set $h^{\prime }(c)=0$ and solve.

$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}\square$

Rolle’s theorem (the special case)

$f(a)=f(b)$: the function starts and ends at the same height. Somewhere in between it must turn around — that turning point has a horizontal tangent ($f^{\prime }(c)=0$).

Hypothesis: $f$ continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$.

Thesis: There exists $c\in (a,b)$ with $f^{\prime }(c)=0$.

Geometric intuition: if you start and end at the same height, you either stayed flat the whole time (constant function, derivative is zero everywhere) or you went up and came back down (or down and back up). At the highest (or lowest) point, the tangent must be horizontal.

Proof

Step 1. $f$ is continuous on $[a,b]$, so by the extreme value theorem (Weierstrass), $f$ attains a maximum $M$ and minimum $m$ somewhere on $[a,b]$.

Step 2. If $M=m$, then $f$ is constant, and $f^{\prime }(x)=0$ for all $x\in (a,b)$. Done.

Step 3. If $M\ne m$, at least one of them differs from $f(a)=f(b)$. That extremum occurs at some interior point $c\in (a,b)$ (not at an endpoint, since both endpoints have the same value).

Step 4. At an interior extremum of a differentiable function, the derivative is zero (Fermat’s theorem: if $c$ is a local max/min and $f^{\prime }(c)$ exists, then $f^{\prime }(c)=0$).

$f^{\prime }(c)=0\square$

Why MVT reduces to Rolle

The MVT proof (above) constructs $h(x)=f(x)-L(x)$ where $L$ is the secant line. This $h$ satisfies $h(a)=h(b)=0$ — exactly Rolle’s hypothesis. So the entire MVT is just “subtract the secant line, apply Rolle’s.”

Limitations

• Existential, not constructive. MVT says $c$ exists but gives no formula for it. You cannot compute $c$ without additional information about $f$.
• Requires differentiability on the entire open interval. If $f$ has even one point where $f^{\prime }$ does not exist (e.g., $|x|$ at $x=0$), MVT does not apply.
• Weaker than the Fundamental Theorem of Calculus. See Connection to the Mean Value Theorem --- FTC gives a complete computation (${\int }_a^b{f}^{\prime }(x)dx=f(b)-f(a)$), while MVT only guarantees one point where the derivative hits the average.

See also

• The Fundamental Theorem of Calculus
• Derivatives, Differentials, and the Chain Rule