Question

Determine all critical points the the function:

I need help on this. I know the derivative is 2x - 16/ sqrt x ... however I'm not sure as to what else

Answer 1

`\bf y=x^2-32√(x)\implies \cfrac{dy}{dx}=2x-32\cdot \cfrac{1}{2}x^{-(1)/(2)}\implies \cfrac{dy}{dx}=2x-\cfrac{32}{2√(x)} \\\\\\ \cfrac{dy}{dx}=2x-\cfrac{16}{√(x)}\implies \cfrac{dy}{dx}=\cfrac{2√(x^3)-16}{√(x)}\\\\ -----------------------------\\\\`


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