Answer
(a). Let $y = (2x)^x$, so that $ln(y) = x ln(2x)$. Then $\frac{1}
{y} y' = 1 + ln(2x)$. Thus $y' = (2x)^x(1 +
ln(2x))$. This quantity is zero when $1+ln(2x) = 0$,
which occurs when $ln(2x) = −1$, or $x = \frac{1}{2e}
≈ .184$.
(b). We have $f(.1) ≈ .851$, $f
(\frac{1}{2e})= e^{−(1/2e)} ≈ .832$,
and $f(1) = 2$. So the absolute minimum is $e^{−(1/2e)}$
and the absolute maximum is $2$.
Work Step by Step
See the graph for explanation.
