Let f(x)=1,,1+x x, 0 x2.

Question

Let $f(x)=\min\{1,\,1+x\sin x\},\ 0\le x\le2\pi$. If $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous, then the ordered pair $(m,n)$ is equal to:

Accepted Answer

Compare $x\sin x$ to $0$: for $x\in(0,\pi)$, $\sin x>0$ so $x\sin x>0\implies1+x\sin x>1\implies f(x)=1$. For $x\in(\pi,2\pi)$, $\sin x<0$ so $x\sin x<0\implies1+x\sin x<1\implies f(x)=1+x\sin x$. $$f(x)=\begin{cases}1, & 0\le x\le\pi\ 1+x\sin x, & \pi

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