Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 6

Rolle's theorem states that a continuous function $g:[u,v]\to \mathbb{R}$ with $u<v$, differentiable on $(u,v)$, and satisfying $g(u)=g(v)$ has $g^{\prime }(c)=0$ for some $c\in (u,v)$.

Fix $x>0$ and set $r=(\sqrt{1+x}-1)/x$. The functionis continuous and differentiable as required, and $g(0)=g(x)=0$. By Rolle's theorem, some $c\in (0,x)$ satisfiesSince $c>0$, this gives $r<1/2$. Multiplying by $x>0$ and adding one proves the strict square root bound

Apply Rolle's theorem on each $[x_i,x_{i+1}]$. This gives $n$ ordered zeros of the derivative $f^{\prime }$ strictly between the original $n+1$ zeros of $f$. Apply Rolle's theorem to $f^{\prime }$ between each consecutive pair of its zeros to obtain $n-1$ ordered zeros of $f^{\prime \prime }$. Continue in this way.

To justify the iteration, for $j<n$, the derivative $f^{(j)}$ is differentiable throughout $(a,b)$ and hence continuous there by differentiability implies continuity. Every new closed interval lies strictly inside $(a,b)$, so Rolle's theorem applies at each step. After $j$ steps there are $n+1-j$ zeros of $f^{(j)}$. At $j=n$ this givesThis is the distinct-zero case of the Rolle root count with multiplicities.

Use the endpoint continuity of $f$ from part b, and interpret the displayed endpoint derivatives as one-sided endpoint values. Starting with $f(a)=f(b)=0$, Rolle's theorem gives $c_1\in (a,b)$ with $f^{\prime }(c_1)=0$. Since also $f^{\prime }(b)=0$, apply Rolle's theorem to $f^{\prime }$ on $[c_1,b]$ to obtain $c_2\in (c_1,b)$ with $f^{\prime \prime }(c_2)=0$. Repeating givesFor $j<n$, the lower derivative $f^{(j)}$ is continuous in the interior by differentiability implies continuity, and at $b$ by the existence of its next one-sided derivative. At the final step, the stipulated continuity of $f^{(n)}$ on $[a,b]$ and its differentiability on $(a,b)$ give exactly the hypotheses of Rolle's theorem on $[c_n,b]$. ThereforeThe endpoint regularity matters: merely assigning arbitrary values to $f$ at $a,b$ while prescribing only the interior derivatives would not justify the first application of Rolle's theorem.

The extreme value theorem gives finite real numbersSince these values are attained, the set of strict upper bounds in the definition of $M$ is precisely $(\beta ,\infty )$, and the set of strict lower bounds in the definition of $m$ is precisely $(-\infty ,\alpha )$. They are nonempty and bounded on the required sides. By the definitions of infimum and supremum,

A maximum or minimum can occur at an endpoint. The rule that a derivative vanishes at an extremum requires an interior extremum and differentiability there. The hypotheses give no two-sided derivative at the endpoints, and even an existing one-sided derivative need not vanish there.

For example, $f(x)=x$ on $[0,1]$ has $m=0$ at $x_m=0$ and $M=1$ at $x_M=1$, but $f^{\prime }(x)=1$ at every interior point; its endpoint one-sided derivatives are also one. Its graph meets both horizontal bounding lines without being tangent to either. Thus the claimed conclusion about two zeros of the derivative fails.

There is a second logical issue: the argument never establishes two distinct extremizing points. For a constant function, the maximum and minimum coincide and may be represented by the same point. The constant function does have many stationary points, but that needs its own argument; it does not follow from counting the same selected point twice. In Rolle's theorem, equal endpoint values ensure an interior extremum when the function is nonconstant, which is the missing mechanism here.