Calculus: Early Transcendentals (2nd Edition)

(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 $2$.