To find the maximum value, we can use the second derivative test where, given f(x), the second derivative < 0, makes it a maximum value.
x(x+1)2=x(x2+2x+1)=x3+2x2+x
=3x2+4x+1=0
Solving, we have x= or −1.
=6x+4
When x=,=2>0
When x=−1,=−2<0
At maximum value of x being -1, y=−1(−1+1)2=0