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

We have, f(x) = |x - 1| f(1) = |1 - 1| = 0 Rf'(1) = h 0 h = h 0 h = h 0 h = h 0 h = 1 and Lf'(1) = h 0 - h = h 0 - h = h 0 h = h 0 - h = - 1 Thus, Rf'(1) Lf'(1) This shows that f(x) is not differentiable at x = 1.

Continuity and Differentiability — Class 12 Maths Solution

ncert exercise SA NCERT Ex.5.2 ,Q.9,Page 166

Question

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

Step-by-step Solution

We have, f(x) $=$|x$-$1|
$f(1) = |1 - 1| = 0$

$Rf'(1) = \mathop {\lim }\limits_{h \to 0} \cfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to 0} \cfrac{{|1 + h - 1| - 0}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \cfrac{{|h|}}{h} = \mathop {\lim }\limits_{h \to 0} \cfrac{h}{h} = 1$

and $Lf'(1) = \mathop {\lim }\limits_{h \to 0} \cfrac{{f(1 - h) - f(1)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \cfrac{{|1 - h - 1| - 0}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \cfrac{{| - h|}}{h} = \mathop {\lim }\limits_{h \to 0} \cfrac{h}{{ - h}} = - 1$

Thus,$Rf'(1) \ne Lf'(1)$

This shows that f(x) is not differentiable at $x = 1.$

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