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An interior point where the derivative equals the average rate of change, proved by tilting the graph by its own chord.
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By the end of this lesson you will be able to prove the interior extremum lemma, Rolle's theorem and the mean value theorem in that order, say which hypothesis each standard counterexample removes, and use the theorem the way it is actually used — converting a bound on the derivative into a bound on the change in the function, without ever identifying the point it produces.
The extreme value theorem, which produces a point where a maximum is taken, and the derivative as a limit of difference quotients. Rolle's theorem is those two facts placed side by side, and everything else in this unit follows from Rolle.
A point is interior to an interval when it is not one of the ends. The chord joins the two end points of the graph, and its slope is the average rate of change across the interval. Rolle's theorem is the case in which that chord is horizontal.
The interior extremum lemma. If $f$ has a local maximum at an interior point $p$ and is differentiable there, then $f'(p) = 0$. Why: for $h > 0$ the quotient $\frac{f(p+h)-f(p)}{h}$ is at most $0$, and for $h < 0$ it is at least $0$; both tend to $f'(p)$, so $f'(p) \le 0 \le f'(p)$.
Rolle's theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then $f'(p) = 0$ for some $p \in (a,b)$.
Proof. The extreme value theorem gives a largest and a smallest value, both attained. If both equal the common end value then $f$ is constant and every interior point works. Otherwise one of them is attained at an interior point, and the lemma applies there.
The mean value theorem. Under the same hypotheses without $f(a) = f(b)$, some interior $p$ has $$f'(p) = \frac{f(b) - f(a)}{b - a}.$$
Proof. Subtract the chord: apply Rolle to $g(x) = f(x) - f(a) - \frac{f(b)-f(a)}{b-a}(x-a)$, which vanishes at both ends.
The tilt is the whole idea. Rolle is the special case, the general case is the special case of the tilted function, and Every hypothesis in this subject is doing work somewhere in the proof. The way to know a theorem is to know which example breaks when a hypothesis is dropped.
Another way: picture
Draw the chord joining the two end points of the graph. Slide a line parallel to the chord up or down until it last touches the curve: at that point the tangent is parallel to the chord. Rolle is the same picture with a horizontal chord, which is why tilting reduces one to the other.
Another way: steps
To use the mean value theorem:
Step 4 is the point. The theorem almost never produces a $p$ anyone cares about; it produces an equation in which $p$ can be estimated away.
Suppose $|f'(t)| \le K$ everywhere. For any $x$ and $y$, the theorem gives $f(x) - f(y) = f'(p)(x-y)$ for some $p$ between them, so $$|f(x) - f(y)| \le K|x - y|.$$ A pointwise bound on the derivative has become a global bound on the function. That conversion is the theorem's purpose, and almost every inequality in analysis is this argument:
Notice what is not needed: nothing about $p$ beyond its existence, and nothing about $f'$ being continuous.
| Hypothesis on an interval | Conclusion | Proof |
|---|---|---|
| $f' = 0$ throughout | $f$ is constant | $f(x) - f(y) = f'(p)(x-y) = 0$ |
| $f' > 0$ throughout | $f$ is strictly increasing | the same identity, with a positive factor |
| $f' = g'$ throughout | $f - g$ is constant | apply the first row to $f - g$ |
Each needs the domain to be an interval. On the domain $(-\infty, 0) \cup (0, \infty)$ the function $\operatorname{sgn}(x)$ has zero derivative everywhere and is not constant, because no single interval joins the two pieces and no chord can be drawn across the gap.
The third row is what makes the fundamental theorem of calculus's second half work, and it is why an antiderivative is determined up to a constant on an interval and not in general.
Ignoring one bad point. Differentiability is asked for at every interior point. The absolute value on a symmetric interval satisfies everything else and defeats the conclusion.
Expecting a usable $p$. The theorem is an existence statement. Trying to compute $p$ in a general argument misses the point, which is that $p$ appears inside a quantity that gets bounded.
Applying the corollaries off an interval. Zero derivative gives constant on an interval only. A domain in two pieces allows a different constant on each.
Confusing it with the extreme value theorem. That one produces a point where the value is largest; this one produces a point where the derivative matches an average. They are different theorems, and this one is proved from that one.
Show $|\sin x - \sin y| \le |x - y|$ for all real $x, y$.
A global claim about a function.
Sine is continuous and differentiable everywhere, so the theorem gives $p$ between $x$ and $y$ with $\sin x - \sin y = \cos(p)(x - y)$.
The theorem, applied on the interval between them.
Since $|\cos p| \le 1$, the inequality follows. The point $p$ was never identified and never needed to be.
Bound the derivative, discard the point.
Show $x^3 - 3x + c$ has at most three real roots, for any constant $c$.
A claim about how many solutions exist.
Between two consecutive roots Rolle gives a point where the derivative $3x^2 - 3$ vanishes.
Roots of $f$ force roots of $f'$.
The derivative has exactly two roots, so $f$ has at most three. Rolle used backwards counts roots, which is one of its standard jobs.
One more root than the derivative has.
Let $f(t) = e^t - 1 - t$, so $f(0) = 0$ and the claim is that $f \ge 0$ everywhere.
Turn the inequality into a claim about one function.
For $x > 0$ the theorem gives $p$ in $(0, x)$ with $f(x) - f(0) = f'(p)x = (e^p - 1)x$, and $e^p > 1$, so $f(x) > 0$.
Apply the theorem on the interval between the two points.
For $x < 0$ the same identity holds with $p$ in $(x, 0)$, where $e^p - 1 < 0$ and $x < 0$, so the product is again positive. Both signs need the same argument with the signs tracked, which is why an inequality proof of this kind is usually two cases rather than one.
A function is continuous on the closed interval from $0$ to $4$, differentiable inside it, and takes the same value at both ends. Put the steps of the proof that its derivative vanishes somewhere inside into order.
Number the steps in order (write the number in the box):
For $f(x) = x^2$ on the closed interval from $0$ to $2$, the mean value theorem gives a point $p$ inside with $f'(p)$ equal to the average rate of change. What is $p$?
Answer:
Each example below defeats the conclusion of Rolle's theorem. Match it to the hypothesis it removes.
| differentiability at every interior point | continuity on the closed interval | equal values at the two ends | a closed interval, so that extreme values are attained | |
|---|---|---|---|---|
| $|x|$ on the closed interval from $-2$ to $2$ | ||||
| the function equal to $x$ below $2$ and to $0$ at $2$, on the closed interval from $0$ to $2$ | ||||
| $x$ on the closed interval from $0$ to $2$ | ||||
| $x$ on the interval from $0$ to $2$ with both ends excluded |
Build the proof of the mean value theorem from Rolle's theorem.
This task has no paper form; do it on a device.
Here is an argument applying the mean value theorem to $f(x) = |x|$ on the interval from $-8$ to $8$. Mark the one step that is not permitted.
This task has no paper form; do it on a device.
A differentiable $f$ satisfies $|f'(t)| \le 6$ for every $t$. What does the mean value theorem give about $|f(x) - f(y)|$?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A function is continuous on the closed interval from $0$ to $2$, differentiable inside it, and takes the same value at both ends. Put the steps of the proof that its derivative vanishes somewhere inside into order.
Number the steps in order (write the number in the box):
You can prove Rolle's theorem from the extreme value theorem, derive the mean value theorem by tilting, and use a derivative bound to bound a function. Say in your own words why the point the theorem produces never needs to be found. Next: what the theorem settles about a function's shape.
10. Your turn: show that $e^x \ge 1 + x$ for every real $x$, step 3