Exercises — Mean Value Theorem — proof, Rolle's theorem

What "check the hypotheses" means: go down the list of three requirements and tick each. Why bother before solving? Rolle only promises a flat tangent when all three hold; skip the check and any $c$ you compute could be a fluke of the algebra, not a guarantee of the theorem.

1. Continuous on $[2,4]$? $f$ is a polynomial. Polynomials are continuous everywhere, so yes. ✔ (See Continuity.)
2. Differentiable on $(2,4)$? Polynomials are differentiable everywhere, so yes. ✔ (See Differentiability.)
3. Equal endpoints, $f(2)=f(4)$? $f(2)=4-12+8=0,f(4)=16-24+8=0.$ Both equal $0$, so $f(2)=f(4)$. ✔ All three hold — Rolle guarantees a flat tangent inside $(2,4)$, even though we haven't located it.

The MVT's target number is the secant slope $\frac{g(b)-g(a)}{b-a}$ — total rise over total run. Why this number? MVT says the derivative equals the average rate of change somewhere; that average rate is exactly this secant slope, so computing it tells us in advance what value $g^{\prime }(c)$ is promised to hit. $\frac{g(9)-g(0)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9}=\frac{3-0}{9}=\frac{1}{3}.$ So MVT promises some $c\in (0,9)$ with $g^{\prime }(c)=\frac{1}{3}$. (Note $g$ is continuous on $[0,9]$ and differentiable on the open interval $(0,9)$ — the derivative $\frac{1}{2\sqrt{x}}$ blows up at $x=0$, but $0$ is an endpoint, not required to be differentiable.)

Step 1 — target (secant slope). Why: MVT says $f^{\prime }(c)$ equals the average rate of change over $[-1,2]$, so we first compute that average — it is the number the derivative must hit. $\frac{f(2)-f(-1)}{2-(-1)}=\frac{4-1}{3}=1.$ Step 2 — set derivative equal. Why set $f^{\prime }(c)=$ secant slope? Because that equation is literally the MVT conclusion; solving it locates the promised point. $f^{\prime }(x)=2x$, so solve $2c=1$: $c=\frac{1}{2}.$ Step 3 — check it's inside. Why: MVT only guarantees $c$ in the open interval, so a valid answer must satisfy $c\in (-1,2)$. Is $\frac{1}{2}\in (-1,2)$? Yes. ✔ Exactly one $c$ here.

Hypotheses: polynomial ⇒ continuous and differentiable; $f(-1)=-1+1=0$, $f(1)=1-1=0$, so $f(-1)=f(1)$. ✔ Rolle applies. Find $c$: Why set $f^{\prime }(c)=0$ here (not a secant slope)? Rolle is the special case of MVT with equal endpoints, so the secant slope is $0$; the promised point is where the tangent is flat, i.e. $f^{\prime }(c)=0$. $f^{\prime }(x)=3x^2-1$, so $3c^2-1=0\Rightarrow c^2=\frac{1}{3}\Rightarrow c=\pm \frac{1}{\sqrt{3}}\approx \pm \mathrm{0.577.}$ Both $+\frac{1}{\sqrt{3}}$ and $-\frac{1}{\sqrt{3}}$ lie in $(-1,1)$. This is a live example of the "at least one, maybe more" rule — here there are two flat tangents: the cubic dips to a valley near $x=-0.577$ and rises to a hump near $x=+0.577$, and at each turning point the tangent is horizontal, matching the slope-$0$ secant that joins the equal-height endpoints.

Step 1 — secant slope. Why the natural log's endpoints: $\ln 1=0$, $\ln e=1$, and this average rate is the value MVT forces $f^{\prime }(c)$ to equal. $\frac{\ln e-\ln 1}{e-1}=\frac{1-0}{e-1}=\frac{1}{e-1}.$ Step 2 — derivative. Why set $\frac{1}{c}$ equal to that? Solving $f^{\prime }(c)=$ secant slope is the MVT conclusion applied to $\ln$. $f^{\prime }(x)=\frac{1}{x}$, so $\frac{1}{c}=\frac{1}{e-1}$ gives $c=e-1\approx \mathrm{1.718.}$ Step 3 — inside? Why: only an interior $c$ counts. $1<e-1<e$ since $e-1\approx 1.718$ and $e\approx 2.718$. ✔

Endpoints: $f(0)=|0-2|=2$, $f(4)=|4-2|=2$. Equal. ✔ Continuity: absolute value is continuous everywhere. ✔ Differentiability on $(0,4)$? Why this is the decisive check: Rolle's guarantee is voided the instant even one hypothesis breaks, so we hunt for a break. The graph is a "V" with its corner at $x=2\in (0,4)$. At a corner the left slope is $-1$ and the right slope is $+1$ — they disagree, so $f^{\prime }(2)$ does not exist. Hypothesis 2 fails. Conclusion: Rolle does not apply, and indeed $f^{\prime }(x)=\pm 1$ never equals $0$. No such $c$ exists. Moral: all other boxes ticked is not enough — one broken hypothesis voids the guarantee.

The flaw is not in MVT — it's that MVT was never allowed to run. Why the check matters: MVT's conclusion is only promised when its hypotheses hold, so before trusting any equation we must verify them. MVT demands $f$ be continuous on the whole closed interval $[-1,1]$. But $\frac{1}{x}$ has a vertical asymptote at $x=0$, which lies inside $[-1,1]$: the function isn't even defined there, let alone continuous. So hypothesis 1 fails and the theorem makes no promise. The "contradiction" $c^2=-1$ is exactly what you'd expect when you apply a theorem outside its domain — the algebra correctly reports that no interior tangent has slope $1$, because the graph jumps from $-\infty$ to $+\infty$ across the gap. Fix: first scan $[a,b]$ for any point where $f$ is undefined or discontinuous. If one exists, stop — MVT is inapplicable.

Goal: show $f(x)=f(y)$ for any two inputs $x<y$, which is exactly what "constant" means. Apply MVT on $[x,y]$ (continuous + differentiable everywhere, so hypotheses hold). Why MVT here? It is the one tool that converts the difference $f(y)-f(x)$ into a single derivative value we already control. There is a $c\in (x,y)$ with $f^{\prime }(c)=\frac{f(y)-f(x)}{y-x}.$ But $f^{\prime }(c)=0$ by assumption. Why this finishes it: the left side is forced to $0$, so the right side (the average change) is $0$, meaning no net change between any two points. So $\frac{f(y)-f(x)}{y-x}=0\Rightarrow f(y)-f(x)=0\Rightarrow f(y)=f(x).$ Since $x,y$ were arbitrary, $f$ takes the same value everywhere — it's constant. $■$ This is the backbone of "zero derivative ⇒ flat function," the tool behind Increasing and Decreasing Functions.

Let $f(x)=x^3+3x-1$. Existence (at least one root) — Intermediate Value Theorem. The Intermediate Value Theorem (IVT) says: if $f$ is continuous on $[a,b]$ and takes values of opposite sign at the endpoints, then $f$ hits every value in between — in particular it must equal $0$ somewhere in $(a,b)$. Why we need it: it manufactures a root out of a sign change. Check its hypotheses: $f$ is a polynomial, so continuous on $[0,1]$ ✔, and $f(0)=-1<0,f(1)=1+3-1=3>0,$ opposite signs ✔. Therefore IVT gives at least one root in $(0,1)$. Uniqueness (no second root) — by contradiction using Rolle. Why Rolle: two roots would be two equal-height points ($f=0$ at both), exactly Rolle's setup, forcing a flat tangent we can then rule out. Suppose there were two roots $a<b$, so $f(a)=f(b)=0$. Then $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and has equal endpoint values — Rolle applies, giving some $c$ with $f^{\prime }(c)=0$. But $f^{\prime }(x)=3x^2+3=3(x^2+1)\ge 3>0\text{for all }x.$ The derivative is never zero, contradicting $f^{\prime }(c)=0$. So two roots is impossible. Conclusion: exactly one real root. $■$

Idea: MVT turns a difference of function values into a single derivative value we can bound. Why that helps: a difference like $\arctan b-\arctan a$ is hard to estimate directly, but one derivative value $\frac{1}{1+c^2}$ is easy to bound. Let $f(x)=\arctan x$, continuous and differentiable on all of $\mathbb{R}$. Case $a\ne b$: apply MVT on the interval with endpoints $a,b$. There is a $c$ between them with $\frac{\arctan b-\arctan a}{b-a}=f^{\prime }(c)=\frac{1}{1+c^2}.$ Take absolute values: $\frac{|\arctan b-\arctan a|}{|b-a|}=\frac{1}{1+c^2}.$ Why the bound now drops out: $c^2\ge 0$, so $1+c^2\ge 1$, hence $\frac{1}{1+c^2}\le 1$. Therefore $|\arctan b-\arctan a|\le |b-a|.$ Case $a=b$: both sides are $0$, so the inequality holds trivially. $■$

Apply MVT on $[0,4]$: Why MVT: it links the unknown endpoint value $f(4)$ to a derivative value we have a ceiling on. Some $c\in (0,4)$ has $f^{\prime }(c)=\frac{f(4)-f(0)}{4-0}=\frac{f(4)-1}{4}.$ Use the bound. Why this pins down the maximum: since every derivative value is $\le 3$, in particular $f^{\prime }(c)\le 3$, which caps the average change and hence $f(4)$: $\frac{f(4)-1}{4}\le 3\Rightarrow f(4)-1\le 12\Rightarrow f(4)\le 13.$ Largest possible value: $f(4)=13$, achieved by the straight line $f(x)=1+3x$ (which has $f^{\prime }\equiv 3$, hitting the bound everywhere). $■$

Step 0 — check the hypotheses, especially $g^{\prime }\ne 0$ on $(1,2)$. Why: Cauchy's MVT requires $g^{\prime }(x)\ne 0$ throughout the open interval, otherwise the ratio $\frac{f^{\prime }(c)}{g^{\prime }(c)}$ could be undefined and $g(2)-g(1)$ could be $0$. Both $f,g$ are polynomials, so continuous on $[1,2]$ and differentiable on $(1,2)$ ✔. And $g^{\prime }(x)=3x^2$, so for $x\in (1,2)$ we have $3x^2\ge 3>0$: $g^{\prime }$ never vanishes. ✔ (Consistently, $g(2)-g(1)=8-1=7\ne 0$.) Cauchy's MVT is now licensed. Step 1 — right-hand side (the ratio of total changes). Why: this is the target value the derivative-ratio must equal. $\frac{f(2)-f(1)}{g(2)-g(1)}=\frac{4-1}{8-1}=\frac{3}{7}.$ Step 2 — left-hand side (ratio of derivatives). Why differentiate first: the left side of Cauchy's conclusion is $f^{\prime }(c)/g^{\prime }(c)$, a ratio of derivatives, not of original values. $f^{\prime }(x)=2x$, $g^{\prime }(x)=3x^2$, so $\frac{f^{\prime }(c)}{g^{\prime }(c)}=\frac{2c}{3c^2}=\frac{2}{3c}.$ Step 3 — set equal and solve. Why: equating the two sides is the Cauchy conclusion; solving it locates $c$. $\frac{2}{3c}=\frac{3}{7}\Rightarrow 14=9c\Rightarrow c=\frac{14}{9}\approx \mathrm{1.556.}$ Inside? $1<\frac{14}{9}<2$. ✔ (Ordinary MVT is the special case $g(x)=x$, where $g^{\prime }=1$ and the ratio becomes the plain secant slope.)

Step 1 — control the slope $f^{\prime }$ using the bound on $f^{\prime \prime }$. Why apply MVT to $f^{\prime }$: we know a ceiling on $f^{\prime \prime }$ (the derivative of $f^{\prime }$), and MVT is exactly the bridge from a derivative bound to the function itself — here the "function" is $f^{\prime }$. For any $x\in (0,1]$, apply MVT to $f^{\prime }$ on $[0,x]$: some $t\in (0,x)$ has $f^{\prime \prime }(t)=\frac{f^{\prime }(x)-f^{\prime }(0)}{x-0}=\frac{f^{\prime }(x)}{x}(\text{using }{f}^{\prime }(0)=0).$ Rearranged, $f^{\prime }(x)=x{f}^{\prime \prime }(t)$, so $|f^{\prime }(x)|=|x||f^{\prime \prime }(t)|\le |x|\cdot 2=2|x|=2x(0<x\le 1).$ Why this is progress: we started knowing only $|f^{\prime \prime }|\le 2$ and now have a bound on $f^{\prime }$ at every point of $[0,1]$. Step 2 — accumulate that slope bound over $[0,1]$ to bound $f(1)$. Why integrate rather than use one more MVT: a single MVT would only give $|f(1)|=|f^{\prime }(s)|\le 2s<2$ for one point $s$, which is too weak. Integrating captures that $f^{\prime }$ starts at $0$ and grows gradually, halving the estimate. Since $f(0)=0$, $f(1)=f(1)-f(0)={\int }_0^1{f}^{\prime }(x)dx,$ so taking absolute values and using $|f^{\prime }(x)|\le 2x$, $|f(1)|=\left|{\int }_0^1{f}^{\prime }(x)dx\right|\le {\int }_0^1|f^{\prime }(x)|dx\le {\int }_0^{1}2xdx=[x^2{]}_0^1=1.$ Conclusion: $|f(1)|\le 1$. $■$ Why the extra care matters: this "bound the derivative, then accumulate it over the interval" move is precisely the logic of the Taylor remainder — the closer you look at how the derivative grows, the sharper the bound on the function.

Recognition (read the hypotheses) ::: check continuity, differentiability, and — for Rolle — equal endpoints. Application (find $c$) ::: set $f^{\prime }(c)$ equal to the secant slope, solve, confirm $c\in (a,b)$. Analysis (why it fails) ::: locate the broken hypothesis — a corner, an asymptote, a jump. Synthesis (prove things) ::: MVT converts differences into single derivative values you can bound. Mastery (generalise) ::: Cauchy's MVT ratios two functions (needs $g^{\prime }\ne 0$); accumulating a derivative bound over the interval seeds Taylor's theorem.