Answer to Question #258992 in Calculus for Pankaj

Question

Is the function f : R ➡R, defined by f(x) = 1-|x| is differentiable at x=1

Answer 1

$\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0^-}\dfrac{1-|1+h|-1+|1|}{h}$

$=\lim\limits_{h\to0^-}\dfrac{-h}{h}=-1$

$\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0^+}\dfrac{1-|1+h|-1+|1|}{h}$

$=\lim\limits_{h\to0^+}\dfrac{-h}{h}=-1$

$\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=-1=\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}$

Then

$\lim\limits_{h\to0}\dfrac{f(1+h)-f(1)}{h}=-1$

Therefore the function $f(x) = 1-|x|$ is differentiable at $x=1$ and $f'(1)=-1.$