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What is the minimum value of the function $y = x^{x}$ ?

$\frac{1}{\pi}^\frac{1}{\pi}$
$0.692$
$\frac{1}{e}^\frac{1}{e}$
$\ln{\pi}$

4 1 3 2

1 solution

Ahaan Rungta
Dec 16, 2013

$y = x^x \implies y' = x^x \cdot (1 + \ln x) = 0 \implies \ln x = -1 \implies x = \dfrac {1}{e}$ .

At $x = \dfrac {1}{e}$ , we have $y = \boxed {\dfrac {1}{e}^{\tfrac{1}{e}}},$ which is choice $\boxed{3}$ .

Hi Ahaan! I think only equating the first derivative to 0 may not always give the point of minima but always gives point of extrema. You can say that at $x=\frac {1}{e}$ , $f(x)$ attains its minima by the first derivative test.

Jit Ganguly - 7 years, 5 months ago

Indeed. To complete it, you need to show that $y''(e^{-1})<0$ .

Sreejato Bhattacharya - 7 years, 5 months ago

I actually did this when doing the problem but forgot to mention it here. My apologies!

Ahaan Rungta - 7 years, 5 months ago

@Ahaan Rungta – That's okay. :)

For the note, it is easy to see that $f(x)=x^x$ is monotonically increasing for $x>1$ , and hence has no global maxima.

Sreejato Bhattacharya - 7 years, 5 months ago

x x x to the x x x

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