2
$\begingroup$

Yes this is homework.

For which values of $a$ is the function $f(x)=e^{-a \sqrt x}$ with $\mathbf dom f = \mathbf R_+$ convex? The possible answers are:

  1. $a \le 0$
  2. $a \ge 0$
  3. $-1 \le a \le 1 $
  4. $a \le -1$

My original thought was that all are convex because for any $a \in \mathbf R, e^{ax}$ is convex. Also, $e^{g(x)}$ is convex if $g$ is convex and I thought that $g(x)=-a \sqrt x$ might be convex (but I don't know the mechanics of how to show this). Finally, I plotted each answer with $x$ some positive scalar across the range of $a$ and they $looked$ convex. Two of the plots are downward sloping and two are upward.

How do I determine if $f(x)=e^{-a \sqrt x}$ with $\mathbf dom f = \mathbf R_+$ is convex?

$\endgroup$
  • $\begingroup$ If $a<0, -a\sqrt{x}$ is not convex. Remember a differentialble function is convex if its graph lies on or above its tangent lines. $\endgroup$ – David Peterson Jan 28 '14 at 21:47
  • $\begingroup$ So given $e^{g(x)}$ is convex if $g$ is convex, does it mean that anywhere that $-a \sqrt x $ is not convex then $e^{-a \sqrt x}$ is not convex? $\endgroup$ – strimp099 Jan 28 '14 at 22:27
  • $\begingroup$ No, the contrapositive would be: if $e^{g(x)}$ is not convex, then $g(x)$ is not convex. My comment was toward your statement "I thought that $g(x)=-a\sqrt{x}$ might be convex." $\endgroup$ – David Peterson Jan 28 '14 at 22:35
  • $\begingroup$ I see, thanks .. $\endgroup$ – strimp099 Jan 28 '14 at 22:41
2
$\begingroup$

Hint: Just compute the second derivative:

$$f'(x) = f(x) \left(-\frac 1 2 a x^{-1/2}\right)$$

and

$$f''(x) = f(x) \left(\frac 1 4 a^2 x^{-1}\right) + f(x) \left(\frac 1 4 a x^{-3/2}\right) = \frac{f(x)}{4 x^{3/2}} \left(a^2 x^{1/2}+a \right)$$

by the chain rule. Now consider values of $a$ to make this positive or negative, after noting that $f$ and $x^{-1}$, $x^{-3/2}$ are certainly positive.

$\endgroup$
  • $\begingroup$ So within the ranges provided for $a$, any value of $\bigtriangledown^2 f(x) $ at the $a$ must be $\succeq 0$ to be convex, correct? In this case $-1 \le a \le 1$ produces negative values so is not convex at those values of $a$... $\endgroup$ – strimp099 Jan 28 '14 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.