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The theorem the rest of the course is proved from, where it comes from, and what it is actually used for.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Rolle's theorem and the mean value theorem with their hypotheses, say how each follows from the extreme value theorem and Fermat's theorem, find the promised point for a given function and interval without assuming it is the midpoint, judge whether a proposed use of a named theorem actually satisfies its hypotheses, and use the theorem as a tool — turning a bound on a derivative into a bound on a change, a sign into monotonicity, and a zero derivative into constancy.
Rolle's theorem. If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there is a $c \in (a,b)$ with $f'(c) = 0$.
The mean value theorem. If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there is a $c \in (a, b)$ with $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$
In words: somewhere strictly inside, the instantaneous rate of change equals the average rate across the whole interval. If a car covers $120$ miles in two hours, at some instant its speedometer read exactly $60$.
The two are the same theorem. Rolle is the case $f(a) = f(b)$; conversely, applying Rolle to $$g(x) = f(x) - \left[f(a) + \frac{f(b)-f(a)}{b-a}(x - a)\right]$$ — the function minus its own secant line, which vanishes at both ends — proves the general case. And Rolle itself follows from the extreme value theorem plus Fermat's theorem: a continuous function on a closed interval attains an extremum, and an interior extremum of a differentiable function is stationary.
So the chain is: completeness → extreme value theorem → Fermat → Rolle → mean value theorem, and from the mean value theorem comes nearly everything left in this course.
A theorem is its hypotheses. Quoting the conclusion of one whose hypotheses you have not checked is not a proof of anything — and in this course it is the single most common way a correct-looking argument turns out to be wrong. Both hypotheses here are load-bearing, and each has a standard counterexample: $|x|$ on $[-1,1]$ is continuous and not differentiable at $0$, and no $c$ has $f'(c) = 0$.
Another way: picture
A curve with the secant line joining its endpoints drawn in. Now slide a line parallel to that secant up until it last touches the curve: where it touches, the tangent is parallel to the secant, and that touching point is $c$. The picture also shows why the theorem promises existence and not location — the touching point could be anywhere inside.
Another way: steps
To use the theorem:
To use it as a tool: write $f(b) - f(a) = f'(c)(b-a)$ and bound $|f'|$.
The mean value theorem is rarely the answer to a question. It is the step inside the proofs of the facts that are.
| Fact | The argument |
|---|---|
| $f' = 0$ on an interval $\Rightarrow$ $f$ constant | $f(q) - f(p) = f'(c)(q-p) = 0$ for any $p, q$ |
| $f' = g'$ $\Rightarrow$ $f - g$ constant | apply the above to $f - g$ |
| $f' > 0$ $\Rightarrow$ $f$ increasing | $f(q) - f(p) = f'(c)(q - p) > 0$ when $q > p$ |
| $\|f'\| \le M$ $\Rightarrow$ $\|f(b)-f(a)\| \le M\|b-a\|$ | take absolute values |
The second row is why the constant of integration exists: two antiderivatives of the same function differ by a constant because their difference has zero derivative. Without the mean value theorem, "$+C$" would be a convention rather than a theorem.
The third row is what makes the sign of $f'$ worth computing at all, and it is the entire justification for the curve-sketching of the next lesson. None of these is obvious, and every one of them is the mean value theorem in two lines.
Quoting it where $f$ is not differentiable. $|x|$ on $[-1, 1]$: the average rate of change is $0$, and $f'$ is never $0$. The failure is at one point and that is enough.
Assuming $c$ is the midpoint. True for a quadratic and false in general. For $x^3$ on $[0, b]$ the point is $b/\sqrt3$.
Assuming $c$ is unique. The theorem says at least one, and a wobbly function can have many.
Using the closed interval for differentiability. Continuity is needed on $[a,b]$, differentiability only on $(a,b)$ — which is what lets the theorem apply to $\sqrt{x}$ on $[0,1]$, where the derivative is infinite at the left end.
Reading it as a statement about averages of $f$. It is about the average rate of change, which is a secant slope. The average value of $f$ itself is a different quantity, and it has its own theorem in the integral unit.
The mean value theorem is usually met as "find the $c$", and that exercise is almost the only thing it is never used for. Nobody outside a calculus course needs to locate the instant a car was doing exactly $60$.
What it is used for is the equation $$f(b) - f(a) = f'(c)(b - a),$$ read left to right as: a change in the function is a derivative times an interval. That converts everything known about $f'$ into a statement about $f$. A bound on $f'$ bounds the change. A sign for $f'$ gives monotonicity. A zero $f'$ gives constancy — and therefore the constant of integration, and therefore the fundamental theorem's second part.
The existence of $c$ is what makes the equation available. That it is unlocatable is not a defect, because nothing that uses the theorem ever needs to locate it.
$f(x) = x^3$ on $[0, 3]$. Average rate of change: $\dfrac{27 - 0}{3} = 9$.
A secant slope.
Solve $f'(c) = 3c^2 = 9$, so $c^2 = 3$ and $c = \sqrt3 \approx 1.732$ — taking the positive root, since $c$ must lie in $(0,3)$.
Roots outside the interval are discarded.
The midpoint is $1.5$, and $\sqrt3$ is not it. The quadratic case is the exception, not the pattern.
Show $|\sin a - \sin b| \le |a - b|$ for all real $a, b$. Apply the mean value theorem to $\sin$ on the interval between them.
$\sin$ is continuous and differentiable everywhere.
$\sin a - \sin b = \cos(c)(a - b)$ for some $c$ between them.
The theorem, written as an equation.
$|\cos c| \le 1$, so $|\sin a - \sin b| \le |a - b|$. A global inequality from a bound on a derivative, in three lines and with no trigonometry at all.
This is the theorem doing what it is for.
Continuous on $[0,4]$, differentiable on $(0,4)$ — the infinite derivative at $0$ is outside the open interval, so the hypotheses hold.
Check the two separately.
Average rate of change: $\dfrac{2 - 0}{4} = \tfrac12$. Solve $\dfrac{1}{2\sqrt c} = \tfrac12$.
$\sqrt c = 1$, so $c = 1$, which is inside $(0,4)$.
Match each theorem condition or conclusion to its role.
| Required closed-interval hypothesis | Required open-interval hypothesis | Equals $f'(c)$ for some $c$ in $(a,b)$ | Guarantees some $c$ with $f'(c)=0$ | |
|---|---|---|---|---|
| Continuous on $[a,b]$ | ||||
| Differentiable on $(a,b)$ | ||||
| $(f(b)-f(a))/(b-a)$ | ||||
| $f(a)=f(b)$ |
Match each theorem condition or situation to the conclusion it licenses.
| Some $c$ has $f'(c)=(f(b)-f(a))/(b-a)$ | Some $c$ has $f'(c)=0$ | $|f(b)-f(a)|\leq M(b-a)$ | The theorem cannot be invoked because continuity fails | |
|---|---|---|---|---|
| Continuous on $[a,b]$ and differentiable on $(a,b)$ | ||||
| Those hypotheses plus $f(a)=f(b)$ | ||||
| $|f'(x)|\leq M$ on $[a,b]$ | ||||
| A jump discontinuity lies inside $[a,b]$ |
For $f(x) = 3x^2$ on $[\,0, 7\,]$, find the $c$ the mean value theorem promises.
Answer:
For $f(x) = x^3$ on $[\,0, 6\,]$, what is the average rate of change across the interval?
Answer:
Is this a legitimate use of the theorem: Rolle's theorem applied to $f(x) = x^2 - 4$ on $[-2, 2]$?
$|f'(x)| \le 9$ everywhere, and $f$ is differentiable. What is the largest $|f(6) - f(0)|$ can be?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = x^2 - 81$ has $f(-9) = f(9) = 0$. Where is the $c$ Rolle's theorem promises?
Answer:
You can find the promised $c$, check hypotheses, and use the theorem to bound a change. Say in your own words why a zero derivative on an interval forces the function to be constant, and why that needs a theorem. Next: monotonicity, concavity and the shape of a graph.
8. Your turn: find the $c$ for $f(x) = \sqrt{x}$ on $[0, 4]$, step 3