Piecewise function and differentiation quotient

Question

We have function $y = f(x) = |2x + 1|$

Can we simply say that $|2x + 1|' = 2$?

No because if we use the chain rule then we get $\left(\left|2x+1\right|\right)'\:=\frac{2\left(2x+1\right)}{\left|2x+1\right|}$

Redefine the same function piecewise without using the absolute value.

$$\left|2x+1\right|= \begin{cases} 2x+1 & x \geq -\frac12 \\ -2x-1 & x \leq -\frac12 \end{cases}$$

Calculate the differential quotient for each of the corresponding regions.

$f(x)=2x+1$

Differential quotient

$$f'(x_0) = \lim_{x→x_0} \frac{f(x) − f(x_0)}{x − x_0}.$$

$f'(x_0) = \lim_{x→x_0} \frac{2x+1 − 2x_0-1}{x − x_0}=2$

$f(x)=-2x-1$

$f'(x_0) = \lim_{x→x_0} \frac{-2x-1 + 2x_0+1}{x − x_0}=-2$

What happens at $x = −\frac12$?

This is where I am confused. There is no x so that would mean that my answer is wrong isn't it?

Answer 1

Note that for the limit of a difference quotient to exist, you must have

$$\lim_{x\rightarrow x_0^{\color{red}{-}}}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\rightarrow x_0^{\color{red}{+}}}\frac{f(x)-f(x_0)}{x-x_0}$$

You do not have that. You have

$$\lim_{x\rightarrow x_0^{\color{red}{-}}}\frac{f(x)-f(x_0)}{x-x_0}=-2\neq 2=\lim_{x\rightarrow x_0^{\color{red}{+}}}\frac{f(x)-f(x_0)}{x-x_0}$$

Therefore your functions is not differentiable at $x=-\frac{1}{2}$

Answer 2

For $x_0> -1/2$ we have $f'(x_0)=2$ and For $x_0< -1/2$ we have $f'(x_0)=-2$.

Furthermore: $f$ is not differentiable at $x_0= -1/2$, since $\lim_{x→x_0+0} \frac{f(x) − f(x_0)}{x − x_0} \ne \lim_{x→x_0-0} \frac{f(x) − f(x_0)}{x − x_0}$.

Answer 3

We have $f(x) = 2x+1$, then $f(-0.5)=0$.

Then we have $$\lim_{x \to x_0^+} \frac{2x+1-0}{x-(-0.5)}=2 \ne -2 = \lim_{x \to x_0^-}\frac{-2x-1-0}{x-(-0.5)}$$

Hence $\lim_{x \to x_0} \frac{2x+1-0}{x-(-0.5)}$ doesn't exist and it is not differetiable at that point.