Let f:(-2,2)R be defined by [f(x)=casesx[x], & -2

Question

Let $f:(-2,2)\to\mathbb{R}$ be defined by
$$f(x)=\begin{cases}x[x], & -2<x<0\ (x-1)[x], & 0\le x<2\end{cases}$$
where $[x]$ denotes the greatest integer function. If $m$ and $n$ respectively are the number of points in $(-2,2)$ at which $y=|f(x)|$ is not continuous and not differentiable, then $m+n$ is equal to \_\_\_\_\_\_\_.

Answer: 4

Accepted Answer

Breaking into sub-intervals where $[x]$ is constant:
$$f(x)=\begin{cases}-2x, & -2<x<-1\ -x, & -1\le x<0\ 0, & 0\le x<1\ x-1, & 1\le x<2\end{cases}$$

Each piece is already $\geq0$ on its interval, so $|f(x)|=f(x)$ except we must check sign/continuity at the joins.

Continuity: at $x=-1$: $f(-1^-)=2,\ f(-1)=1$ --- discontinuous. At $x=0$: $f(0^-)=0=f(0)$ --- continuous. At $x=1$: $f(1^-)=0=f(1)$ --- continuous. So $m=1$.

Differentiability: $f'(x)=-2$ on $(-2,-1)$, $-1$ on $(-1,0)$, $0$ on $(0,1)$, $1$ on $(1,2)$ --- all different at each junction, so $f$ (and hence $|f|$, which coincides with $f$ here) is non-differentiable at $x=-1,0,1$: $n=3$.

$m+n=1+3=4$

Solutions · 1