Find $\alpha \in \mathbb{R}$ s.t. the second derivative ($x=0$) of a function exists.

Question

Find $\alpha \in \mathbb{R}$ s.t. the second derivative ($x=0$) of the following function exists.

$f(x) = \begin{cases} e^{-\frac{1}{x}}, & \text{if $x$}\gt 0 \\[2ex] \sin(x)+\alpha x^2-\log(1+x), & \text{if $x$ } \leq0 \end{cases}$

How to deal with this kind of exercise? Is it enough to use Taylor at second order and then looking for the differentiability of the function?

Answer 1

Taylor at second order means that you have already calculated the second derivative. Just calculate the second derivative on the left and on the right, and make them equal. $$\left(e^{-\frac 1x}\right)''=-\frac{2x-1}{x^4}e^{-\frac 1x}$$ This term goes to $0$ as $x\to 0$. For the other side the second derivative is $$-\sin x+2\alpha+\frac{1}{(x+1)^2}$$ The limit at $0$ is $2\alpha+1$. So $2\alpha+1=0$ should give you the answer.

Answer 2

The function is continuous at $0$, by checking the limits.

The derivative is $$ f'(x)=\begin{cases} \dfrac{e^{-1/x}}{x^2} & x>0 \\[4px] \cos x+2\alpha x-\dfrac{1}{1+x} & x<0 \end{cases} $$ Since the limits of $f'$ for $x\to0$ from the left and from the right are both equal to $0$, we can say that the function is differentiable at $0$ and that the derivative is continuous. It follows from l'Hôpital's theorem.

Now apply the same technique, with the difference that $$ \lim_{x\to0^+}f'(x)=0 $$ (check it) and $$ \lim_{x\to0^-}f'(x)=\lim_{x\to0^-}\left(-\sin x+2\alpha+\frac{1}{(1+x)^2}\right)=2\alpha+1 $$

A sufficient condition for differentiability of $f'$ at $0$ is $\alpha=-1/2$. Is it also necessary? Note that l'Hôpital can be used only one way.

Answer 3

Hint

A function whose right and left derivatives are given in a point, is differentiable at that point if those right and left derivatives are equal.

Based on this, try to differentiate the function twice, each time checking for which values of $a$ are the right and left derivatives in $x=0$ equal.