Question

Determine all critical points the the function:

Answer 1

$\bf y=x^2-32\sqrt{x}\implies \cfrac{dy}{dx}=2x-32\cdot \cfrac{1}{2}x^{-\frac{1}{2}}\implies \cfrac{dy}{dx}=2x-\cfrac{32}{2\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=2x-\cfrac{16}{\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{2\sqrt{x^3}-16}{\sqrt{x}}\\\\ -----------------------------$

$\bf 0=\cfrac{2\sqrt{x^3}-16}{\sqrt{x}}\implies 0=2\sqrt{x^3}-16\implies 16=2\sqrt{x^3} \\\\\\ 8=\sqrt{x^3}\implies 2^3=\sqrt{x^3}\implies (2^3)^2=x^3\implies (2^6)^{\frac{1}{3}}=x \\\\\\ 2^{\frac{2}{1}}=x\implies \boxed{4=x}\\\\ -----------------------------\\\\ \textit{the other critical point at }\sqrt{x}=0\implies \boxed{x=0}$