Question

Let $f\left(x\right)=\left\{\begin{array}{ll}1,& x\le -1\\ \left|x\right|,& -1<x<1\\ 0,& x\ge 1\end{array}\right.$ Then, f is

(a) continuous at x = − 1
(b) differentiable at x = − 1
(c) everywhere continuous
(d) everywhere differentiable

Answer 1

(b) differentiable at x = − 1

$f\left(x\right)=\left\{\begin{array}{ll}1,& x\le -1\\ \left|x\right|,& -1<x<1\\ 0,& x\ge 1\end{array}\right.$

Differentiabilty at x = − 1
(LHD x = − 1)
$$\begin{gathered} \underset{x\to -1^{-}}{lim}\frac{f\left(x\right)-f(-1)}{x+1} \\ =\underset{x\to -1}{lim}\frac{f\left(x\right)-f(-1)}{x+1} \\ =\underset{x\to -1}{lim}\frac{1-1}{-1+1} \\ =0 \end{gathered}$$

(RHD x = − 1)
$$\begin{gathered} =\underset{x\to -1^{+}}{lim}\frac{f\left(x\right)-f(-1)}{x+1} \\ =\underset{x\to -1}{lim}\frac{f\left(x\right)-f(-1)}{x+1} \\ =\underset{x\to -1}{lim}\frac{f\left(x\right)-f(-1)}{x+1} \\ =\underset{x\to -1}{lim}\frac{\left|x\right|-|-1|}{x+1} \\ =\frac{1-1|}{-1+1} \\ =0 \end{gathered}$$