Question

Let $g\left(x\right)=\{\begin{array}{cc}2(x+1),& -\infty <x\le -1\\ \sqrt{1-x^2},& -1<x<1\\ |||x|-1|-1|,& 1\le x<\infty \end{array}.$ Then

• A. $g\left(x\right)$ is non-differentiable at exactly three points
• B. $g\left(x\right)$ is continuous in $(-\infty ,1]$
• C. $g\left(x\right)$ is differentiable in $(-\infty ,-1)$
• D. $g\left(x\right)$ has finite type of discontinuity at $x=1$, but continuous at $x=-1$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $g\left(x\right)$ is non-differentiable at exactly three points
C $g\left(x\right)$ is differentiable in $(-\infty ,-1)$
D $g\left(x\right)$ has finite type of discontinuity at $x=1$, but continuous at $x=-1$



From the above graph, we can verify the alternatives.