FIND THE DERIVATIVES FROM THE LEFT AND RIGHT AT THE GIVEN POINT

For a function y = f(x) defined in an open interval (a, b) containing the point x₀, the left hand and right hand derivatives of f at x = h are respectively denoted by f'(h^-) and f'(h⁺)

f'(h^-)  =  lim _{h-> 0}^-[f(x + h) - f(x)] / h

f'(h⁺)  =  lim _{h-> 0}⁺[f(x + h) - f(x)] / h

provided the limits exist.

Question 1 :

Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

(i)  f(x)  =  |x - 1|

Solution :

If the function is differentiable, then

f'(1^-)  =  f'(1⁺)

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

=  lim_x->1- [-(x - 1) - 0]/(x - 1)

=  -1

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

=   lim_x->1+ [(x - 1) - 0]/(x - 1)

=  1

Hence the given function is not differentiable at x = 1.

(ii)  f(x)  =  √(1 - x²)

Solution :

If the function is differentiable, then

f'(1^-)  =  f'(1⁺)

f'(1^-)  =  [f(x) - f(1)] / (x - 1)

=   lim_x->1- [√(1 - x²) - 0]/(x - 1)

=   lim_x->1- [√(1 - x²) - 0]/(1 - x)

=   lim_x->1- -√(1 + x) / √(1 - x)

=  -√2 / 0

=  -∞

Hence the given function is not differentiable at x = 1.

Solution :

If the function is differentiable, then

f'(1^-)  =  f'(1⁺)

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

=  lim_x->1- (x - 1)/(x - 1)

=  1

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

=   lim_x->1+ (x² - 1)/(x - 1)

=  lim_x->1+ (x + 1)(x - 1)/(x - 1)

=  lim_x->1+ (x + 1)

=  2

f'(1^-)  =  1 and f'(1⁺)  =  2,  so the given function is not differentiable at x = 1.

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