Question

For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

Answer 1

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$

$\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1$

All this to say that f(x) is differentiable everywhere on its domain, except at the point x = -1.