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

Answers (1)

Given function is 
f (x) = |x-1 | , x \epsilon R
We know that any function is differentiable when both
\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}       and      \lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}   are finite and equal
Required condition for function to be differential at x  = 1 is   

\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}
Now, Left-hand limit of a function at x = 1 is
\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^-}\frac{|h|-0}{h}    
                                                                                                             = \lim_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)
Right-hand limit of a function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^+}\frac{|h|-0}{h}
                                                                                                             =\lim_{h\rightarrow 0^-}\frac{h}{h} = 1
Now, it is clear that 
R.H.L. at x= 1  \neq  L.H.L. at x= 1
Therefore, function f (x) = |x-1 | is not differentiable at x = 1
 

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