For which values of $a$ is $e^{-a\sqrt x}$ convex in $\mathbb R^+$?

Question

I failed the following question in a quiz:

For which values of $a$ the function $e^{-a\sqrt x}$ with $dom = \mathbb{R}^+$ is convex? Check all that apply:

1. $a\leq0$
2. $a\geq0$
3. $-1 \leq a\leq1$
4. $a\leq-1$

I said that in all cases it is convex because when plotting for me it was kind of obvious. However, the correct answer is 2. Why?

Thank you.

Answer 1

Well, why not to differentiate twice and check?

$$f(x):=e^{-a\sqrt x}\;,\;\;f'(x)=-\frac a{2\sqrt x}e^{-a\sqrt x}\;,$$

$$f''(x)=\frac{e^{-a\sqrt x}}{4x}a\left(\frac 1{\sqrt x}+a\right)$$

Now

$$f''(x) \geq 0\iff a\left(\frac1{\sqrt x}+a\right) \geq 0\iff\begin{cases} a \geq 0\;,\;\;\;\text{or} \\{}\\a<0\;\;\text{and}\;\;\frac1{\sqrt x}<-a\end{cases}$$