simplify f(-t^2 +3) when f(x)= 4x[sqr(3-x)]

Question

simplify f(-t² +3) when f(x)= 4x[sqr(3-x)]

Answer 1

The question is asking you to substitute the "x" in the equation with "(-t²+3).

This gives you:

f(-t²+3)=4*(-t²+3)*√(3-(-t²+3))

Simplify by using distribution.

So,

4*(-t²+3)=-4t²+(4*3)=-4t²+12

And,

3-(-t²+3)=3+t²-3=(3-3)+t²=0+t²=t²

REMEMBER that a negative times a negative equals a positive (-1*-t²=t²)

Now you have:

f(-t²+3)=(-4t²+12)(√t²)

You know that:

(√t²)=t

So,

f(-t²+3)=(-4t²+12)t

Distribute the "t" through the parenthesis

ANSWER:

f(-t²+3)=-4t³+12t