Question

If f(x) = |x + 100| + x², test whether f’(–100) exists.

Answer 1

f(x) = |x + 100| + x²

First let us find the left limit of f(x) at x = – 100

When x < – 100 ,

f(x) = – (x + 100) + x²

f(– 100) = – (– 100 + 100) + (– 100)²

f(– 100) = 100²

`f"'"(- 100^-) =  lim_(x -> - 100^-) (f(x) - f(- 100))/(x - (- 100)`

= `lim_(x -> -10^-) (-(x + 100) + x^2 - 100^2)/(x + 100)`

= `lim_(x -> -100^-) [(-(x + 100))/(x + 100) + (x^2 - 100^2)/(x + 100)]`

= `lim_(x -> -100^-) [- 1 + ((x + 100)(x - 100))/(x + 100)]`

= `lim_(x -> -100^-) [- 1 + x - 100]`

= – 1 – 100 – 100

f'(– 100) = – 201  ........(1)

Next let us find the right limit of f(x) at x = – 100

when x > – 100

f(x) = x + 100 + x²

f(– 100) = – 100 + 100 + (– 100)²

f(– 100) = 100² 

`f"'"(- 100^+) =  lim_(x -> - 100^+) (f(x) - f(- 100))/(x - (- 100))`

= `lim_(x -> - 100^+) (x + 100 + x^2 - 100^2)/(x + 100)`

f'(– 100⁺) = – 199  ........(2)

From equation (1) and (2), we get

f’(– 100^–) ≠ f'(– 100⁺)

∴ f’(x) does not exist at x = – 100

Hence, f'(– 100) does not exist