# What are the local extrema of f(x)= e^xln1^x?

Nov 12, 2015

${1}^{x} = 1$ for all $x$, so $\ln {1}^{1} = \ln 1 = 0$
Therefore, $f \left(x\right) = {e}^{x} \ln {1}^{x} = {e}^{x} \cdot 0 = 0$ for all $x$.
$f$ is a constant. The minimum and maximum of $f$ are both $0$.