Define $f: \mathbb{R} \to \mathbb{R}$ by $ \ f(x)=\begin{cases} x^2, & \text{if } x<0, \\ x^2+2x, & \text{if } x \geq 0. \end{cases}$
Then which of the following statements are correct ?
(i) $ \ f''(x)=2 \ $ for all $ \ x \in \mathbb{R} \ $
(ii) $ f''(0) \ $ does not exist
(iii) $ f'(x) \ $ exists for each $ \ x \neq 0 \ $
(iv) $ \ f'(0) \ $ does not exist
Answer:
At first try to check whether $ \ f'(x) \ $ exists or not at $ \ x=0 \ $ .
Then,
$$ R \ f'(0)= \lim_{h \to 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^{+}} \frac{h^2+2h}{h}=2 $$
$$ L \ f'(0)= \lim_{h \to 0^{-} } \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^{+}} \frac{h^2}{h}=0 $$
Thus $ L \ f'(0) \neq R \ f'(0) \ $
Thus $ \ f \ $ is not differentiable at $ \ x=0 \ $
This means $ \ f'(0) \ $ does not exists.
Thus option $(iv)$ is correct .
Also option $(iii)$ is also correct.
Therefore $(iii)$ and $(iv)$ are correct.
I need confirmation of my work.
Is there any help?
Your argument showing that $(iv)$ is right is valid.
That $(iii)$ is right is shown by ordinary differentiation formulas, and can also be shown by finding limits as you did in the argument that shows that $(iv)$ is right.
$f''(0)$ cannot exist because $f'(0)$ does not exist.