Question

Let $f\left(x\right)=\{\begin{array}{c}2x+3,-3\le x<-2\\ x+1,-2\le x<0\\ x+2,0\le x\le 1\end{array}$. At what points the function is/are not differentiable in the interval $(-3,1)$

• A. $-2$
• B. $0$
• C. $1$
• D. $1/2$

Answer 1

Answer: (A), (B)

$-2$
$0$

Given that $f\left(x\right)=\{\begin{array}{c}2x+3,-3\le x<-2\\ x+1,-2\le x<0\\ x+2,0\le x\le 1\end{array}$, we have
$f^{\prime }\left(x\right)=\{\begin{array}{c}2,-3\le x<-2\\ 1,-2\le x<0\\ 1,0\le x\le 1\end{array}$
Thus, $f^{\prime }\left(2^{-}\right)\ne f^{\prime }\left(2^{+}\right)$ implies the function is not differentiable at $x=-2.$
However, the function is discontinuous at $x=0$ and thus, it is also not differentiable at $x=0.$