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!


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

  • $\begingroup$ 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

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