Chapter 7: Transcendental Functions - Practice Exercises - Page 440: 115

$-\frac{1}{2}$ is minimum $0$ is maximum

$\frac{dy}{dx}$ = $x(\frac{2}{2x})+\ln2x-1$ = $\ln2x$ $\frac{dy}{dx}$ = $0$ so $\ln2x$ = $0$ $2x$ = $e^{0}$ $x$ = $\frac{1}{2}$ if $x$ = $\frac{1}{2}$ then $y$ = $\frac{1}{2}\ln1-\frac{1}{2}$ $y$ = $-\frac{1}{2}$ is minimum if $x$ = $\frac{e}{2}$ then $y$ = $\frac{e}{2}\ln{e}-\frac{e}{2}$ $y$ = $0$ is maximum