Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 10F (Analysis)

Differentiability at $x_0$ meansHence $f(x)-f(x_0)=(x-x_0)(f^{\prime }(x_0)+o(1))\to 0$ as $x\to x_0$. Therefore every differentiable function is continuous at each point where it is differentiable.

The mean value theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f(b)-f(a)=f^{\prime }(c)(b-a)$ for some $c\in (a,b)$.

For $F(t)=\cos (e^{-t})$ and $t\ge 0$,The mean value theorem gives $|\cos (e^{-x})-\cos (e^{-y})|\le |x-y|$.

For the second inequality put $t=\sqrt{1+x}\ge 1$. It becomes $2\log t\le t-t^{-1}$. The difference $G(t)=t-t^{-1}-2\log t$ satisfies $G(1)=0$ andThus $\log (1+x)\le x/\sqrt{1+x}$.

For $x\ne 0$, $f(x)=|x{|}^x=e^{x\log |x|}$, so $f^{\prime }(x)=|x{|}^x(\log |x|+1)$. Although $x\log |x|\to 0$ makes $f$ continuous at zero, $(f(x)-1)/x\sim \log |x|\to -\infty$, so it is not differentiable there.

Since the cosine function is even, $g(x)=\cos |x|=\cos x$, so it is differentiable everywhere with $g^{\prime }(x)=-\sin x$. Finally, $h(x)=x|x|$ is differentiable everywhere, including at zero, and $h^{\prime }(x)=2|x|$. Therefore

At a nonzero rational $p/q$ in lowest terms, irrational points approach $p/q$ with function value zero, whereas $f(p/q)=1/q>0$, so $f$ is discontinuous. At zero, $|p/q|\ge 1/q$ for nonzero $p$, hence $0\le f(x)\le |x|$ and continuity follows.

At an irrational $x$, given $\epsilon >0$, only finitely many rationals in a bounded neighbourhood have reduced denominator $q\le 1/\epsilon$. Choose a neighbourhood avoiding those points; throughout it, $f<\epsilon$. Consequently the continuity set is $(\mathbb{R}\setminus \mathbb{Q})\cup \{0\}$. This is a variant of the Thomae function.