Question

1 ⩽x ⩽ 5, for √(x³ + 36 )

Answer 1

$\bf ~\hspace{10em}1\le x\le 5~\hspace{5em}\sqrt{x^3+36} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt{x^3+36}\implies \sqrt{x^{2+1}6^2}\implies \sqrt{x^2x6^2}\implies \sqrt{(6x)^2x}\implies 6x\sqrt{x} \\\\\\ \stackrel{\textit{x = 4}}{6(4)\sqrt{4}}\implies 24\sqrt{2^2}\implies 24\cdot 2\implies 48$

that's how I read it.... to get some value between 1 and 5, namely 4, to make the expression a rational, well, 48 can be expressed as 48/1.