HOW TO CHECK DIFFERENTIABILITY OF A FUNCTION AT A POINT

The function is differentiable from the left and right. As in the case of the existence of limits of a function at x₀, it follows that

exists if and only if both

exist and f' (x₀-)  =   f' (x₀+) 

Hence 

if and only if f' (x₀-)  =   f' (x₀+) . If any one of the condition fails then f'(x) is not differentiable at x_0.

Question 1 :

(i)  f(x) = |x| + |x - 1| at x = 0, 1

Solution :

f(x) = |x| + |x - 1|

Check if the given function is continuous at x = 0.

If x < 0, then f(x) = -x - (x - 1)

f(x)  =  -x - x + 1

  =  -2x + 1

If x > 0 and x < 1, then f(x) = x - (x - 1)

f(x)  =  x - x + 1

  =  1

If x > 1, then f(x) = x + (x - 1)

f(x)  =  x + x + 1

  =  2x + 1

f'(0^-)  =  lim _x->0- [(f(x) - f(0)) / (x - 0)]

  =  lim _x->0- [(-2x + 1) - 1] / x

  =  lim _x->0- -2x / x

  =  lim _x->0- -2

  =  -2  -----(1)

f'(0⁺)  =  lim _x->0+ [(f(x) - f(0)) / (x - 0)]

  =  lim _x->0+ [1 - 1] / x

  =  lim _x->0- 0 / x

  =  0/0  -----(2)

f'(0^-)  ≠ f'(0⁺) 

f'(1^-)  =  lim _x->1- [(f(x) - f(1)) / (x - 1)]

  =  lim _x->1- [1 - 1] / (x - 1)

  =  lim _x->1- 0 / (x-1)

  =  0 -----(1)

f'(1⁺)  =  lim _x->1+ [(f(x) - f(1)) / (x - 1)]

  =  lim _x->1+ [2x + 1 - 3] / (x - 1)

  =  lim _x->1+ (2x - 2) / (x - 1)

  =  lim _x->1+ 2(x - 1) / (x - 1)

  =  2 -----(2)

f'(1^-)  ≠ f'(1⁺) 

Hence the given function is not differentiable at the given points.

(ii)  f(x)  =  sin |x| at x = 0

Solution :

If x < 0, then f(x) = sin (-x)  

f(x)  =  - sin x 

If x > 0, then f(x) =  sin x

f (x)  =  sin x

f'(0^-)  =  lim _x->0- [(f(x) - f(0)) / (x - 0)]

  =  lim _x->0- [(-sin x) - 0] / x

  =  lim _x->0- -(sin x/x)

  =  -1  -----(1)

f'(0⁺)  =  lim _x->0+ [(f(x) - f(0)) / (x - 0)]

  =  lim _x->0+ [(sin x) - 0] / x

  =  lim _x->0- (sin x/x)

  =  1  -----(2)

f'(0^-)  ≠ f'(0⁺) 

Hence the given function is not differentiable at x = 0.

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