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<0\end{cases}`

or more simply,

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

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<-1\end{cases}`

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.