1
$\begingroup$

I am doing some review questions for an upcoming final and stumbled upon the following:

$ f(x) = \left\{ \begin{array}{lr} 1 \ : if \ x = 0\\ x^\sqrt{x} \ : if \ 0 < x \le 2 \end{array} \right. $

Find the absolute minimum value of $f(x)$ on $[0,2]$

(A) $e^{-1/e}$

(B) $e^{-2/e}$

(C) $e^{-1/2e}$

(D) $e^{-2\sqrt{e}}$

(E) $e^{-\sqrt{e}}$

So far, I know I need to use Rolle's Therom to try to find a value c in $[0,2]$ that is the absolute minimum but I don't know how to utilize it since this is a piecewise function. Any hints or help is greatly appreciated! Cheers!

$\endgroup$
0
$\begingroup$

Find the critical numbers of $f(x)$ in $0<x<2$. Compare the value of $f$ there with the value of $f$ at the endpoints, $x=0$ and $x=2$. The smallest of these numbers is the absolute min. This is due to the Extreme Value Theorem for continuous functions on a closed interval.

The critical number is at $x=1/e^2$ and $f(0)=1$, $f(2)=2^\sqrt{2}\approx 2.67$, $f(1/e^2)=e^{-2/e}\approx 0.48$. Thus, the absolute min is $e^{-2/e}$.

Here's a graph:

enter image description here

$\endgroup$
  • $\begingroup$ On a test (or in general) it is important to note that this is true because the function is continuous on $[0,2]$. $\endgroup$ – Eric Dec 7 '13 at 18:44
  • $\begingroup$ Try the first derivative test then to analyze the general behavior of the graph. $\endgroup$ – Eric Dec 7 '13 at 18:49
  • $\begingroup$ @wonggr: Do you mean the graph of $f(x)$? Or do you mean the numerical values of $f(0)$, $f(2)$, and $f(1/e^2)$? $\endgroup$ – JohnD Dec 7 '13 at 18:49
  • $\begingroup$ I was going to ask for the numerical value but I took the natural logarithm of both sides (i.e. $y = f(1/e^2)$) and got the answer! Thank you all for the help! $\endgroup$ – wonggr Dec 7 '13 at 18:58

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.