Past exam of the mathematical methods II course of the University of Oxford / 2019 / 3 / Solution

The real logarithm restricts the domain to $x>0$. Differentiating givesSince $x$ cannot vanish in the domain, a stationary point requires $\ln (3x)=-1/2$. The logarithm is strictly increasing, so this equation has exactly one solution:The second derivative is $f^{\prime \prime }(x)=2\ln (3x)+3$, giving $f^{\prime \prime }(x_{\ast })=2>0$. In fact $f^{\prime }$ is negative below $x_{\ast }$ and positive above it, so the point is the unique global minimum on $x>0$. The boundary value $x=0$ is not an additional stationary point, since the original function is not defined there.