Math Problem Solver

$x^{x}$ has a local minimum $e^{- \frac{1}{e}} \approx 0.6922$ at $x = e^{-1}$.

$x^{x}$ has no local maxima.

$x^{x}$ has a global minimum $e^{- \frac{1}{e}} \approx 0.6922$ at $x = e^{-1}$.

$x^{x}$ has no global maxima.

To find local extrema of a function, use the second derivative test. A critical point of a single-variable function $f$ at a point $c$ is a local minimum if $f''(c) > 0$ and a local maximum if $f''(c) < 0$.

$x^{x}$ has a critical point at $x = e^{-1}$. Use the second derivative test on each critical point to determine if any of these points are local extrema.
Take the second derivative. $$f''(x) = x^{x} \left(\left(\log{\left(x \right)} + 1\right)^{2} + \frac{1}{x}\right)$$

Evaluate at $x = e^{-1}$. $$\begin{aligned}f''(x) &= e^{-1}^{e^{-1}} \left(\left(1 + \log{\left(e^{-1} \right)}\right)^{2} + e^{-1}^{-1}\right) \\ &= e^{1 - e^{-1}}\end{aligned}$$$e^{1 - e^{-1}} > 0$ so $x=e^{-1}$ is a local minimum. Evaluate the function at $x = e^{-1}$ to find the value of the minimum. $$\begin{aligned}f(e^{-1})&=e^{-1}^{e^{-1}} \\ &=e^{- \frac{1}{e}}\end{aligned}$$$f(x)$ has a local minimum $e^{- \frac{1}{e}}$ at $x = e^{-1}$.