Answer
$f(x)$ is not differentiable at the origin.
Work Step by Step
$f(x)=2x-1$ for $x\ge0$ and $f(x)=x^2+2x+7$ for $x\lt0$
- The right-hand derivative of $f(x)$ at the origin:
As $h\to0^+$, $f(h)=2h-1$ and $f(0)=2\times0-1=-1$. Therefore, $$\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{2h-1-(-1)}{h}=\lim_{h\to0^+}\frac{2h}{h}=\lim_{h\to0^+}2=2$$
- The left-hand derivative of $f(x)$ at the origin:
As $h\to0^-$, $f(h)=h^2+2h+7$ and $f(0)=-1$. Therefore, $$\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{h^2+2h+7-(-1)}{h}=\lim_{h\to0^-}\frac{h^2+2h+8}{h}$$
As $h\to0^-$, $(h^2+2h+8)\to8$. So $\lim_{h\to0^-}\frac{h^2+2h+8}{h}=-\infty$
Since the left-hand derivative differs with the right-hand one, we conclude that $f(x)$ is not differentiable at the origin..
