If F(x)=3-x^2 and g(x)=1+2x find (and simplify) the defining equation of the composition function f/circle g.?

Answer 1

F(g(x)= 3- {(1+2x)(1+2x)
= 3 - {(1+2x+2x+4x^2)}
= 3 - {(1+4x+4X^2)}
= 3 - 1 - 4X - 4X^2
= 2-4X-4x^2
= -4x^2 - 4x +2
Now, using quadratic formula ( also know as almighty formula),
we have: {4 (plus or minus) Square root of 48}/ -8
which is also
= 4 (+-) 4 square root 3/ -8
= 4{(1 (+-) square root of 3)}/ -8
= (1+- square-root 3)/ -2.
you can put the negative on the denominator on top
like this: -(1+- square root 3)/ 2

Answer 2

If F(x)=3-x^2 and g(x)=1+2x find (and simplify) the defining equation of the composition function f/circle g.

F(g(x)) = 3-(1+2x)^2
= 3-(1+4x+4x^2)
= 3-1-4x-4x^2
= 2-4x-4x^2

Answer 3

Well composite functions work so that (f o g) (x) = f(g(x))

It can be a little confusing at first, since the x's in both equations are not the same.

(fog) (x) = f(g(x))=f(1+2x)=3-(1+2x)^2

Answer 4

f o g = f(g(x))
f o g=3-(g(x))^2
=3-(1+2x)^2
=3-(1+4x+4x^2)
=-4x^2-4x+2

Answer 5

f[g(x)]
=3-[1+2x]^2
=3-1-4x-4x^2
=2-4x-4x^2
=2[1-2x-2x^2]