Answer
The proof is detailed below.
Work Step by Step
- The right-hand derivative of $f(x)$ at $P(1,2)$:
As $x\to1^+$, $f(x)=2x$. Therefore, $$\lim_{h\to1^+}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^+}\frac{2(h+1)-2}{h}=\lim_{h\to1^+}\frac{2h+2-2}{h}$$ $$=\lim_{h\to1^+}\frac{2h}{h}=\lim_{h\to1^+}2=2$$
- The left-hand derivative of $f(x)$ at $P(1,2)$:
As $x\to1^-$, $f(x)=2$. Therefore, $$\lim_{h\to1^-}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^-}\frac{2-2}{h}=\lim_{h\to1^-}0=0$$
Since the left-hand derivative differs with the right-hand one, we conclude that $f(x)$ is not differentiable at $P$.
