Verification of Problems?

Question

F x G: F(x) = Sqrt(2x +3) and G(x) = (x^2) +1

Ok, I did the problems, but it would be really helpful if someone checked them for me:

G x F: G(f(x)) = G(Sqrt(2x +3)) = (Sqrt(2x +3))^2 + 1
the Sqrt and ^2 cancel each other out leaving 2x +3 + 1 = 2x +4
domain should be all real numbers,nothing makes it undefined.

F x F: F(F(Sqrt(2x +3)) = Sqrt(2(Sqrt(2x +3)) +3) this is the final equation but for the domain the inside Sqrt equation has to be > 0 so 2x + 3 > 0... 2x > -3.... x> -3/2 is my domain

G X G: G(G(X^2 +1)) = (X^2 +1)^2 +1 = x^4 + x^2 + x^2 +1 + 1 = x^4 + 2x^2 +2..the domain is all real numbers because nothing makes the equation undefined.

Answer 1

Thanks for showing what you've done!

In G x F, there are two issues. 1) the domain still needs to restrict x to x> -3/2 just as in the F x F part of the problem. 2) Technically, when you cancel the sqrt and the second power, the amount that is left over should never be negative. (For example, sqrt(-2)^2 = sqrt(4) = 2, not -2.) So
(Sqrt(2x +3))^2 + 1 = absolute value of (2x + 3) + 1, however, due to the restriction that x>-3/2, we already know that (2x + 3) >= 0, so the absolute value is unnecessary.


F x F: Sqrt{2[Sqrt(2x +3)] +3} can be simplified a bit more.
Sqrt{2[Sqrt(2x +3)] +3}
= Sqrt{Sqrt 4[(2x +3)] +3}
= Sqrt{Sqrt(8x +12) +3}


G x G looks well done.

I hope this helps!

Answer 2

Looks like you got it.