Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 6 / a

The mean value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, there is $c\in (a,b)$ such thatNo continuity assumption on the derivative is needed.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.

The derivative of $h(x)=f(x)-\sin x$ vanishes. By the constancy theorem on the interval $I$,This need not hold on a disconnected domain. On $I=(-\infty ,0)\cup (0,\infty )$, define $f(x)=\sin x$ for $x<0$ and $f(x)=\sin x+1$ for $x>0$. The function is differentiable throughout $I$ with $f^{\prime }(x)=\cos x$, but no single $C$ works on both components.