Question

Show that the function g(x) = |x + 2| is not differentiable at x = - 2 .​

Answer 1

Given : g(x) = |x + 2|

To Find : Show that  not differentiable at x = - 2 .​

Solution:

differentiable  if  

LHD = RHD

LHD =

Lim (h → 0)  (g(x) -g(x - h) )/h  

=  Lim (h → 0)  (g(-2) -g(-2 - h) )/h  

= Lim (h → 0)  (|-2 + 2|  - (-2 - h+2) )/h

= Lim (h → 0)  (0  - |-h|)/h

|-h| = h

= Lim (h → 0)  (0 - h  )/h

= Lim (h → 0)  -h/h

=  Lim (h → 0)   -1

= - 1

RHD

Lim (h → 0)  (g(x+h) -g(x) )/h  

=  Lim (h → 0)  (g(-2+h) -g(-2) )/h  

= Lim (h → 0)  (|-2 + h +2|  - (-2+2) )/h

= Lim (h → 0)  (|h|   -0)/h

= Lim (h → 0)  (h - 0 )/h

= Lim (h → 0)   h/h

=  Lim (h → 0)    1

=   1

1 ≠ 1

LHD  ≠ RHD

Hence not differentiable at x = - 2

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Answer 2

SOLUTION

TO PROVE

The function g(x) = |x + 2| is not differentiable at x = - 2

EVALUATION

Here the given function is g(x) = |x + 2|

Now we have to check the differentiability of g(x) at x = - 2

Now by definition of modulus function

$\sf |x + 2| = \begin{cases} & \sf{ \: \: \: \: (x + 2) \: \: \: \: when \: x > - 2} \\ \\ & \sf{ - (x + 2) \: \: \: \: \: when \: x \leqslant - 2} \end{cases}$

∴ g(-2) = 0

Left hand derivative at x = - 2

$= \displaystyle \sf{ \lim_{x \to 2 - } \: \frac{g(x) - g( - 2)}{x - ( - 2)} }$

$= \displaystyle \sf{ \lim_{x \to 2 - } \: \frac{ - (x +2 ) }{ \: \: \: \: (x + 2)} }$

$= - 1$

Right hand derivative at x = - 2

$= \displaystyle \sf{ \lim_{x \to 2 + } \: \frac{g(x) - g( - 2)}{x - ( - 2)} }$

$= \displaystyle \sf{ \lim_{x \to 2 + } \: \frac{ (x +2 ) }{ (x + 2)} }$

$= 1$

Hence at x = - 2

Left hand derivative ≠ Right hand derivative

∴ g(x) = |x + 2| is not differentiable at x = - 2

Hence proved

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