For the function f(x)=e^|x|-|x|, xR, consider the following statements: Statement I: f is …

Question

For the function $f(x)=e^{\sin|x|}-|x|,\ x\in\mathbb{R}$, consider the following statements:

Statement I: $f$ is differentiable for all $x\in\mathbb{R}$.

Statement II: $f$ is increasing in $\left(-\pi,-\dfrac{\pi}{2}\right)$.

Accepted Answer

Since $|x|$ (and hence $\sin|x|$) has a corner at $x=0$, $f(x)=e^{\sin|x|}-|x|$ is not differentiable at $x=0$. So Statement I is false. For $x\in\left(-\pi,-\dfrac{\pi}{2}\right)$, $x<0$ so $|x|=-x$ and $\sin|x|=\sin(-x)=-\sin x$, giving $f(x)=e^{-\sin x}+x$. $$f'(x)=-e^{-\sin x}\cos x+1$$ On this interval $\cos x<0$, so $-e^{-\sin x}\cos x>0$, hence $f'(x)>0+1>0$ --- wait, more precisely $f'(x)=1+(-\cos x)e^{-\sin x}>0$ since both terms are positive. So $f$ is increasing on this interval. Statement II is true.

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