simplify f(-t^{2} +3) when f(x)= 4x[sqr(3-x)]
The question is asking you to substitute the "x" in the equation with "(-t^{2}+3).
This gives you:
f(-t^{2}+3)=4*(-t^{2}+3)*√(3-(-t^{2}+3))
Simplify by using distribution.
So,
4*(-t^{2}+3)=-4t^{2}+(4*3)=-4t^{2}+12
And,
3-(-t^{2}+3)=3+t^{2}-3=(3-3)+t^{2}=0+t^{2}=t^{2}
REMEMBER that a negative times a negative equals a positive (-1*-t^{2}=t^{2})
Now you have:
f(-t^{2}+3)=(-4t^{2}+12)(√t^{2})
You know that:
(√t^{2})=t
So,
f(-t^{2}+3)=(-4t^{2}+12)t
Distribute the "t" through the parenthesis
ANSWER:
f(-t^{2}+3)=-4t^{3}+12t