Question #258912

Is the function f : R ➡R, defined by f(x) = 1-|x| is differentiable in its domain.

1
Expert's answer
2021-11-01T16:15:59-0400

Let us check whether the function f:RRf : \R \to\R , defined by f(x)=1xf(x) = 1-|x| is differentiable in its domain.

Since limx0+f(x)f(0)x0=limx0+1x1x=limx0+xx=limx0+xx=1\lim\limits_{x\to 0+}\frac{f(x)-f(0)}{x-0} =\lim\limits_{x\to 0+}\frac{1-|x|-1}{x} =\lim\limits_{x\to 0+}\frac{-|x|}{x} =\lim\limits_{x\to 0+}\frac{-x}{x}=-1 and

limx0f(x)f(0)x0=limx01x1x=limx0xx=limx0(x)x=limx0xx=11,\lim\limits_{x\to 0-}\frac{f(x)-f(0)}{x-0} =\lim\limits_{x\to 0-}\frac{1-|x|-1}{x} =\lim\limits_{x\to 0-}\frac{-|x|}{x} =\lim\limits_{x\to 0-}\frac{-(-x)}{x} =\lim\limits_{x\to 0-}\frac{x}{x}=1\ne -1,

we conclude that the function ff is not differentiable at the point x=0x=0 , and hence it is not differentiable in its domain.


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