Question

PLEASE HELP! See attachment:

Answer 1

$\bf f(x)=x^2e^{-2x}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=2xe^{-2x}+x^2\cdot -2e^{-2x}\implies \cfrac{dy}{dx}=xe^{-2x}(2-2x) \\\\\\ \cfrac{dy}{dx}=\cfrac{x(2-2x)}{e^{2x}}\implies 0=\cfrac{x(2-2x)}{e^{2x}} \\\\\\ 0=x(2-2x)\implies \begin{cases} 0=x\\ 0=2-2x\implies 2x=2\implies x=1 \end{cases}$

now, bear in mind, that zeroing out the denominator, also gives critical points, usually asymptotic points, where the derivative is undefined, now, in this case, the denominator is never zero, so we don't get any from the denominator, just from the numerator, and are 0 and 1

now check the picture below

running a first-derivative test on it, those are the values on those regions

you get a negative, regardless of what it might be, what matters is the sign
you get a positive, and then a negative

so, f(x) goes down, then up then down

now, you can see, there's on relative minimum and a relative maximum