Question

Where is the second derivative of y = 4xex equal to 0?

Answer 1

$\bf y=4xe^x\implies \cfrac{dy}{dx}=4e^x+4xe^x\implies \cfrac{d^2y}{dx^2}=8e^x+4xe^x \\\\\\ 0=8e^x+4xe^x\implies 0=4e^x(2+x)\implies 0=2+x\implies -2=x$