The function $f(x)=8{\log }_{e}x-x^2+3$ attains its global minimum over the interval $[1,e]$ at $x=\dots$

$f^{\prime }(x)=\frac{8}{x}-2x$.Set $f^{\prime }(x)=0⟹8=2x^2⟹x^2=4⟹x=2$ (Since $x\in [1,e]$ and $e\approx 2.71$).Test values at endpoints and critical point:$f(1)=8(0)-1+3=2$$f(2)=8\ln 2-4+3=8(0.693)-1\approx 4.54$$f(e)=8(1)-e^2+3=11-7.39=3.61$ Comparing $\{2,4.54,3.61\}$, the global minimum is $2$, which occurs at $x=1$.