Eliminate a positive parameter and retain closed endpoints

The requested segment of $y=\tfrac12\ln x$ increases and is concave down from $(2,\tfrac12\ln2)$ to $(5,\tfrac12\ln5)$. Both endpoints are included. Its parameter interval is $\sqrt2\le t\le\sqrt5$.

$y=\ln t$ gives $t=e^y$. Then $x=(e^y)^2=e^{2y}$, so $2y=\ln x$ and $y=\tfrac12\ln x$, $x>0$ for the full curve.

Since $t>0$, $x=t^2$ gives $t=\sqrt x$, not a choice of either sign. Thus $y=\ln\sqrt x=\tfrac12\ln x$, $x>0$. The sketch in (i) uses only $2\le x\le5$.