Can you determine the largest number?

Let $f(x) = x^\frac 1x$ . We note that $2^\frac 12 = 4^\frac 14$ and that $f(x)$ is increasing at $x=2$ but decreasing at $x=4$ , then $f(3) = \boxed{3^\frac 13}$ must be the largest number.

Define $f(x)=x^{\frac{1}{x}}$ on the positive reals. This can also be written as $f(x)=e^{\frac{\log x}{x}}$ . Differentiating, we have

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

Now, $\frac{f(x)}{x^2}$ is clearly positive for all positive $x$ ; so for this function to have a turning point, we must have $1-\log x=0$ or in other words $x=e$ .

Returning to the integers, the maximum must occur at either $n=2$ or $n=3$ (the two nearest integers to $e$ ), and it's easy to check that $\boxed{3^{\frac{1}{3}}}$ is the required maximum. (It's also easy to check that this is a maximum, not a minimum; the easiest way is to consider $f(1)$ or $f(4)$ ).