Let f:RR be defined by f(x)=cases3(1-|x|2), & |x|20, & |x|2cases.

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be defined by $f(x)=\begin{cases}3\left(1-\dfrac{|x|}{2}ight), & |x|\le2\0, & |x|>2\end{cases}$. Let $g:\mathbb{R}	o\mathbb{R}$ be given by $g(x)=f(x+2)-f(x-2)$. If $n$ and $m$ denote the number of points in $\mathbb{R}$ where $g$ is not continuous and not differentiable respectively, then $n+m$ is equal to ..........

Answer: 4

Accepted Answer

$$f(x)=\begin{cases}0, & x<-2\\frac32(1+x), & -2\le x<0\\frac32(1-x), & 0\le x<2\0, & x>2\end{cases}$$

Shifting and combining, $g(x)=f(x+2)+f(x-2)$ (equivalently) is piecewise-linear, continuous everywhere ($n=0$) but has corners at $x=-4,-2,2,4$, giving $m=4$. So $n+m=4$.

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