The best way to answer this question is to graph the function.
You will immediately see:,
>f blows up negatively at x = 0
>f is asymptotic to y = 0 from below as x goes to - infinity
>x = 4 is a zero where f crosses from negative to positive and then is asymptotic to y=0 from above as x goes to positive infinity
f''(x)= (x3-2x2+8x)/x4
and as long as x is not equal to 0 this is (x2-2x+8)/x3
The numerator has no real zeros.
However, there appears to be a local maximum near x=8, but I am unable to find it analytically; I will think about the problem and get back to you if I figure it out.