F ` ( x ) = ( x² )` · e^(5x) + x² · ( e^(5x) )` =
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )
First find the critical points of <em>f</em> :



so the point (1, 0) is the only critical point, at which we have

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

Find the critical points of <em>g</em> :



where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at




So <em>f</em> has a minimum of -7 and a maximum of 299.
I’m pretty sure there are infinite solutions as the system of equations are equivalent to each other.