Unit 07 · Analysis 1A

The Mean Value Theorem

The theorem that turns local information about $f'$ into global information about $f$. The proof chain — EVT → Fermat → Rolle → MVT — is one of the most examinable sequences in the course, and every link is short.

By the end you can

Prove Fermat's theorem, Rolle's theorem, and the MVT in order.
Deduce constancy and monotonicity from the sign of $f'$.
Prove inequalities with the MVT.
State the Cauchy MVT and Darboux's theorem.

Why it matters & where it comes from

Why it exists

It is the rigorous bridge from a derivative at one point to behaviour across a whole interval — why $f'=0$ everywhere forces $f$ constant, a fact that has no proof without it.

In the real world

Average-speed cameras use it literally; numerical analysis uses it for error bounds; differential equations use it for uniqueness of solutions.

Story & open questions

Named partly for Rolle (1691), himself an early critic of calculus; Cauchy gave the modern statement and the generalized (Cauchy) Mean Value Theorem behind L'Hôpital's rule.

Step 1 — Fermat's theorem

Theorem Fermat

If $f$ has a local maximum or minimum at an interior point $c$ and $f'(c)$ exists, then $f'(c)=0$.

Proof (local max case)

For small $h>0$: $f(c+h)-f(c)\le0$, so the quotient $\dfrac{f(c+h)-f(c)}{h}\le0$, and letting $h\to0^+$ gives $f'(c)\le0$.

For small $h<0$ the quotient is $\ge0$, giving $f'(c)\ge0$. Both hold, so $f'(c)=0$. ∎

Points where $f'=0$ or $f'$ does not exist are the critical points — the only interior candidates for extrema.

Step 2 — Rolle, Step 3 — MVT

Theorem Rolle

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof

By the EVT (Unit 5), $f$ attains a max and a min on $[a,b]$.

If both happen at endpoints, then max $=$ min (as $f(a)=f(b)$), so $f$ is constant and $f'\equiv0$. Otherwise some extremum is interior, and Fermat gives $f'(c)=0$ there. ∎

Theorem MVT

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ for some $c\in(a,b)$.

Proof

Tilt the picture flat: let $g(x)=f(x)-\left[f(a)+\dfrac{f(b)-f(a)}{b-a}(x-a)\right]$, the gap between $f$ and its secant line.

$g$ inherits continuity and differentiability, and $g(a)=g(b)=0$; Rolle gives $g'(c)=0$, which rearranges to the MVT. ∎

Consequences

Corollaries on an interval

If $f'=0$ everywhere, $f$ is constant. If $f'>0$ everywhere, $f$ is strictly increasing (similarly $f'<0$, decreasing). If $f'=g'$ everywhere, then $f-g$ is constant.

Proof (constancy)

For any $x_1<x_2$ in the interval, the MVT on $[x_1,x_2]$ gives $f(x_2)-f(x_1)=f'(c)(x_2-x_1)=0$. All values agree. ∎

Example Proving an inequality

$\ln(1+x)<x$ for $x>0$ (using $(\ln t)'=1/t$ from Unit 11):

MVT for $\ln$ on $[1,1+x]$: $\ln(1+x)-\ln1=\tfrac1c\cdot x$ for some $c\in(1,1+x)$.

Since $c>1$, $\tfrac1c<1$, so $\ln(1+x)=\tfrac xc<x$. ∎

Two upgrades worth knowing

Theorem Cauchy MVT

If $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, there is $c\in(a,b)$ with $\big(f(b)-f(a)\big)g'(c)=\big(g(b)-g(a)\big)f'(c)$. (Apply Rolle to $h(x)=f(x)\big(g(b)-g(a)\big)-g(x)\big(f(b)-f(a)\big)$.) This is the engine inside L'Hôpital's rule.

Theorem Darboux

Derivatives have the intermediate value property even when they are not continuous: if $f$ is differentiable on $[a,b]$ and $y$ lies between $f'(a)$ and $f'(b)$, then $f'(c)=y$ for some $c$. So the discontinuous derivative of Unit 6's showpiece still never "jumps."

Common pitfalls

• Check both hypotheses: continuity on the closed interval, differentiability on the open one.
• The MVT gives existence of $c$ — not its value, and not uniqueness.
• "$f'(x_0)>0$ at one point" does not make $f$ increasing near $x_0$; the clean statements are about $f'>0$ on an interval.
• The corollaries need an interval: $f'=0$ on $(0,1)\cup(2,3)$ allows different constants on each piece.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · Constancy

Prove: $f'=0$ on an interval $\Rightarrow$ $f$ constant there.

Solution

For any $x_1<x_2$, the MVT gives $f(x_2)-f(x_1)=f'(c)(x_2-x_1)=0$, so all values are equal.

Practice 2 · Lipschitz bound

Show $|\sin a-\sin b|\le|a-b|$ for all real $a,b$.

Solution

MVT on $\sin$: $\sin a-\sin b=\cos(c)(a-b)$; take absolute values and use $|\cos c|\le1$.

Practice 3 · Inequality

Prove $e^x\ge1+x$ for all $x$ (using $(e^x)'=e^x$ from Unit 11).

Solution

$h(x)=e^x-1-x$ has $h(0)=0$ and $h'(x)=e^x-1$, which is negative for $x<0$ and positive for $x>0$. So $h$ decreases then increases: $x=0$ is a global minimum and $h\ge0$ everywhere.

Practice 4 · Rolle by hand

Verify Rolle's theorem for $f(x)=x^2-4x+3$ on $[1,3]$ and find every guaranteed $c$.

Answer

$f(1)=f(3)=0$, $f$ is a polynomial, and $f'(x)=2x-4$ vanishes exactly at $c=2\in(1,3)$.

Practice 5 · Exactly one root

Show $x^3+3x+1=0$ has exactly one real root.

Solution

$f(-1)=-3<0<1=f(0)$ gives a root (IVT). If there were two, Rolle would give a point with $f'=0$; but $f'(x)=3x^2+3\ge3>0$. So exactly one.

Practice 6 · Bounded slope

Suppose $|f'(x)|\le M$ for all $x$ in an interval. Prove $|f(x)-f(y)|\le M|x-y|$ for all $x,y$ there.

Solution

For $x\ne y$ the MVT gives $f(x)-f(y)=f'(c)(x-y)$; take absolute values. (This is called a Lipschitz bound.)

Practice 7 · Trig inequality

Prove $\tan x>x$ for $x\in\left(0,\tfrac\pi2\right)$.

Solution

$g(x)=\tan x-x$ has $g(0)=0$ and $g'(x)=\sec^2x-1=\tan^2x>0$ on $\left(0,\tfrac\pi2\right)$, so $g$ is strictly increasing from $0$.

Practice 8 · Counting zeros

Prove a polynomial of degree $n$ has at most $n$ real zeros.

Solution

$n+1$ distinct zeros would give, by Rolle between consecutive ones, $n$ zeros of $p'$, then $n-1$ of $p''$, …, finally a zero of the $n$th derivative — which is the nonzero constant $n!\,a_n$. Contradiction.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

Fermat and Rolle. Use extrema to produce zero derivatives under endpoint equality.
MVT. Prove and apply the mean value theorem as the central derivative theorem.
Consequences. Derive constant, monotone, Lipschitz, and inequality results.
Error control. Use derivative bounds to control function-value changes.

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The theorem chain is Fermat, Rolle, MVT, then every derivative-controls-function corollary.

Study routine. In applications, identify the interval endpoints first; MVT always compares two function values over a closed interval.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · MVT statement

State the hypotheses and conclusion of the Mean Value Theorem.

Guide

Continuity on $[a,b]$ and differentiability on $(a,b)$ imply some $c\in(a,b)$ satisfies $f\prime(c)=(f(b)-f(a))/(b-a)$.

Extension 2 · Constant derivative

Prove if $f\prime=0$ on an interval then $f$ is constant.

Guide

For any two points, MVT gives the difference quotient equals $0$, so the function values are equal.

Extension 3 · Bound

If $|f\prime(x)|\le3$ on $[a,b]$, prove $|f(b)-f(a)|\le3|b-a|$.

Guide

Apply MVT and take absolute values.

Extension 4 · Hypothesis failure

Why does Rolle fail for $f(x)=|x|$ on $[-1,1]$?

Guide

The endpoint values match, but $f$ is not differentiable at $0$, and there is no point with derivative $0$ in the differentiable parts.

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