5.1 Using the Mean Value Theorem

5.1 Using the Mean Value Theorem

A function can reveal that something must have happened without telling you exactly where it happened. The Mean Value Theorem proves this for rates of change; the Extreme Value Theorem proves that highest and lowest values exist under the right conditions.

Two existence theorems, two different guarantees

Theorem	Required conditions	Guaranteed conclusion
Mean Value Theorem	$f$ is continuous on $[a,b]$ and differentiable on $(a,b)$	At some $c$ in $(a,b)$, $f'(c)=\dfrac{f(b)-f(a)}{b-a}$
Extreme Value Theorem	$f$ is continuous on the closed interval $[a,b]$	$f$ has at least one absolute minimum and at least one absolute maximum on $[a,b]$

Both are existence theorems: they establish that a value or behavior occurs, but they may not identify its exact location. This distinction matters in applications. A theorem may prove that a driver exceeded a speed limit somewhere on a highway without specifying the precise mile marker.

5.1 The Mean Value Theorem: matching an average rate

The average rate of change of $f$ from $x=a$ to $x=b$ is the slope of the secant line:

$$ \frac{f(b)-f(a)}{b-a}. $$

The Mean Value Theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then at least one interior point $c$ satisfies

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

In plain language, the function’s instantaneous rate equals its overall average rate at least once.

Worked contextual example. A driver travels $120$ miles in $2$ hours. Let $s(t)$ denote position, in miles, $t$ hours after departure. The average velocity is

$$ \frac{s(2)-s(0)}{2-0}=\frac{120}{2}=60\text{ miles per hour}. $$

If $s$ is continuous on $[0,2]$ and differentiable on $(0,2)$, the Mean Value Theorem guarantees some time $c$ with

$$ s'(c)=60. $$

If the posted speed limit is $55$ miles per hour, an average of $60$ does not by itself say when the driver went fast, but the Mean Value Theorem guarantees an instant at which the driver’s velocity was exactly $60$ miles per hour, which is enough to conclude that the driver was speeding at that moment.

The assessed reasoning process is 3.E Provide reasons or rationales for solutions and conclusions. A complete justification names the hypotheses—continuity on $[a,b]$ and differentiability on $(a,b)$—then states the theorem’s conclusion and connects it to the context.

Misconception check — “The average rate is the rate at the midpoint.”
The Mean Value Theorem guarantees a point $c$, but not usually $c=\dfrac{a+b}{2}$. The matching point may occur anywhere inside the interval.

Retrieval check. A function $f$ is continuous on $[1,5]$ and differentiable on $(1,5)$, with $f(1)=3$ and $f(5)=11$. What does the Mean Value Theorem guarantee? Answer: at least one $c$ in $(1,5)$ with $f'(c)=\dfrac{11-3}{5-1}=2$. The theorem does not say where $c$ is, and there may be more than one.