Modulus function 's differentiability

Question

How do I show f(x ) = |x | + | x -1 | is not differentiable at x =0 by using the conventional " limits " method ?

Answer 1

Hint: Near $0$, $x-1<0$ and therefore $f(x)=|x|-(x-1)=|x|-x+1$.

Answer 2

Use the definition of a derivative as a limit $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$. Show that the left limit (negative $h$) differs from the right limit (positive $h$).

Answer 3

Hint: You can write $$f(x)=|x|+|x-1|=\begin{cases}-2x+1 & \text{ if } x < 0\\1& \text{ if } 0 \leq x < 1\\2x-1 & \text{ if } x \geq 1.\end{cases}$$ Now use the limit definition of the derivative near $x=0$. $$f^{'}(0+)=\lim_{h \to 0^+}\frac{f(h)-f(0)}{h} \text{ and } f^{'}(0-)=\lim_{h \to 0^-}\frac{f(h)-f(0)}{h}$$

Answer 4

You can split the Function y = |x| into two functions: See:
1. y = -x from -∞ to 0 AND 2. y = x from 0 to ∞ \ So that would mean that you have two different grades at the same x-Location what is not possible. Try it out on your given function!