Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 11D (Analysis I)

Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

LetThen $h(a)=h(b)=0$. Since $f$ is not linear, $h(c)\ne 0$ for some $c\in (a,b)$. If $h(c)>0$, the mean value theorem on $[a,c]$ gives a point $\xi$ withIf $h(c)<0$, apply it on $[c,b]$ to obtainIn either case $h^{\prime }=f^{\prime }-m$, and therefore

No. On $[-1,1]$, takeThen $f^{\prime }(0)=0$. For every $x_1<0<x_2$,This quadratic expression is strictly positive: it equalsThus no chord with endpoints on opposite sides of $\xi$ has slope $f^{\prime }(\xi )=0$.

SetIts endpoint values are equal. Rolle theorem therefore supplies $\xi \in (a,b)$ withIf $g^{\prime }(\xi )=0$, then (1) and $g(b)\ne g(a)$ force $f^{\prime }(\xi )=0$, contrary to the hypothesis. Hence $g^{\prime }(\xi )\ne 0$, and division in (1) gives

The condition is necessary. On $[0,2\pi ]$, letHere $g(2\pi )-g(0)=\pi /2$ while $f(2\pi )-f(0)=0$, so the endpoint ratio is zero. ButWhere $g^{\prime }(x)\ne 0$, their ratio is $1/(1+\frac{1}{2}\cos x)$ and is never zero. At the remaining points both derivatives vanish, so the derivative ratio is undefined. Thus the conclusion fails when simultaneous zeros are allowed.