Prove that the function f given by: f(x) = |x - 1|, x R is not differentiable at x = 1.

Given: f(x) = |x - 1| f(1) = |1 - 1| = 0 Rf ' (1) = ^ lim _ h 0 f(1 + h)-f(1) h = ^ lim _ h 0 |1 + h - 1|-0 h = ^ lim _ h 0 |h| h = ^ lim _ h 0 h h =1 And Lf ' (1) = ^ lim _ h 0 f(1 - h)-f(1) -h = ^ lim _ h 0 |1 - h - 1|-0 -h = ^ lim _ h 0 |-h| h = ^ lim _ h 0 -h h =-1 Since R f ' (1) L f ' (1) Therefore, f(x) is not differentiable at x =1.

Prove that the function f given by: f(x) = |x - 1|, x R is not di…

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Question

Prove that the function f given by:
$\text{f(x)} = |\text{x} - 1|, \text{x} \in \text{R}$
is not differentiable at x = 1.

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Answer

Given: $\text{f(x)} = |\text{x} - 1|\ \therefore \ \text{f(1)} = |1 - 1| = 0$
$\text{R}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 + h)}-\text{f}(1)}{\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 + \text{h} - 1|-0}{\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|\text{h}|}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{h}}{\text{h}}=1$
$\text{And}\ \text{L}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 - h)}-\text{f}(1)}{-\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 - \text{h} - 1|-0}{-\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|-\text{h}|}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{-\text{h}}{\text{h}}=-1$
Since $\text{R }\text{f}{'}(1)\neq \text{L}{\text{f}}{'}(1)$
Therefore, f(x) is not differentiable at x =1.

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