Question

This function
Is it differentiable at x = 1​ ?

Answer 1

By definition of absolute value:

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

Then the derivative is

`(d|x-1|)/(dx) = \begin{cases} 1 & \text{if } x > 1 \\ -1 & \text{if } x < 1\end{cases}`

but does not exist at
`x=1` because

`\displaystyle \lim_(x\to1^-) f'(x) = -1`

while

`\displaystyle \lim_(x\to1^+) f'(x) = 1`

and these limits are not equal, so the derivative is discontinuous and hence
`|x-1|` is not differentiable at
`x=1`.

Answer 2

It is not differentiable at x=1 since the slope of the tangent line as x -> 1 from the right is 1 while the slope of the tangent line as x->1 from the left is -1