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

## About "Find the Derivatives From the Left and Right at the Given Point"

Find the Derivatives From the Left and Right at the Given Point :

Here we are going to see how to check if the function is differentiable at the given point or not.

How to Find if the Function is Differentiable at the 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 x0.

## How to Know If a Function is Differentiable at a Point - Examples

Question 1 :

Determine whether the following function is differentiable at the indicated values.

(i) f(x) = x | x | at x = 0

Solution :

f(x) = x | x |

If x < 0, then f(x) = x (-x)  =  -x

If x > 0, then f(x) = x (x)  =  x

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

=  lim x->0- (-x2 - 0) / x

=  lim x->0- -x

=  0  -----(1)

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

=  lim x->0+ (x2 - 0) / x

=  lim x->0+ x

=  0 -----(2)

f'(0-)  =  f'(0+

Hence the given function is differentiable at the point x = 0.

(ii)  f(x) = |x2 - 1| at x = 1

Solution :

f(x) = |x2 - 1|

If x < 1, then f(x) = -(x2 - 1)

If x > 1, then  f(x) = (x2 - 1)

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

=  lim x->1- [(-x2 + 1) - (0)] / (x - 1)

=  lim x->1- [-(x2 - 1) / (x - 1)]

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

=  -2  -----(1)

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

=  lim x->1+ [(x2 - 1) - (0)] / (x - 1)

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

=  lim x->1+ (x + 1)

=  2  -----(2)

f'(1-)  ≠ f'(1+

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

(iii)  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.

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