Given the position function, the velocity function is obtained by taking the derivative:
[tex]s(t)=(t+1)(t-3)^3\implies v(t)=(t-3)^3+3(t+1)(t-3)^2[/tex]
The velocity is increasing its own derivative is positive, so we also have to find the acceleration by taking another derivative:
[tex]a(t)=4(t-3)^3+3(t-3)^2+6(t+1)(t-3)[/tex]
To find when [tex]a(t)>0[/tex], we first need to know where [tex]a(t)=0[/tex]:
[tex]4(t-3)^3+3(t-3)^2+6(t+1)(t-3)=(t-3)\bigg(4(t-3)^2+3(t-3)+6(t+1)\bigg)=(t-3)(4t^2-15t+33)=0[/tex]
The quadratic factor is always positive (its discriminant is negative), which leaves one solution at [tex]t=3[/tex]. To either side of [tex]t=3[/tex], we have, for instance,
[tex]a(2)=-12<0[/tex]
[tex]a(4)=36>0[/tex]
which indicates that [tex]v(t)[/tex] is increasing for [tex]t>3[/tex], making the answer A.
