Answer
Concave up for $x\gt9$
Concave down for $0\lt x\lt9$
Point of inflection at $x=9$
Work Step by Step
$y=f(x)=x(x-8\sqrt x)=x^{2}-8x^{\frac{3}{2}}$
$f'(x)=2x-8\times\frac{3}{2}x^{\frac{3}{2}-1}=2x-12x^{\frac{1}{2}}$
$f''(x)=2-12\times\frac{1}{2}x^{-\frac{1}{2}}=2-\frac{6}{\sqrt x}$
The function is concave up when $f''(x)\gt 0$.
$f''(x)\gt0\implies2-\frac{6}{\sqrt x}\gt0$$\implies\sqrt x\gt\frac{6}{2}\implies x\gt9$
The function is concave up when $x\gt9$.
Also, as $f''(x)\lt0$ for $x\in (0,9)$, $f$ is concave downward for $0\lt x\lt9$.
$f''(x)=0\implies x=9$ and as $f''(x)$ changes sign at $x=9$, the point of inflection is at $x=9$.
