Question #258992

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

Expert's answer

lim⁡h→0−f(1+h)−f(1)h=lim⁡h→0−1−∣1+h∣−1+∣1∣h\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0^-}\dfrac{1-|1+h|-1+|1|}{h}

=lim⁡h→0−−hh=−1=\lim\limits_{h\to0^-}\dfrac{-h}{h}=-1



lim⁡h→0+f(1+h)−f(1)h=lim⁡h→0+1−∣1+h∣−1+∣1∣h\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0^+}\dfrac{1-|1+h|-1+|1|}{h}

=lim⁡h→0+−hh=−1=\lim\limits_{h\to0^+}\dfrac{-h}{h}=-1

lim⁡h→0−f(1+h)−f(1)h=−1=lim⁡h→0+f(1+h)−f(1)h\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⁡h→0f(1+h)−f(1)h=−1\lim\limits_{h\to0}\dfrac{f(1+h)-f(1)}{h}=-1

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


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Guyana #340795 on Jan 2024