$\displaystyle F(x)=\sqrt{x}+1$

$\displaystyle G(x)=\sqrt{x+1}$

Prove:

$\displaystyle G(F(x))=F(g(x))$

I'm a little confused as to how I would prove x=0 and without eliminating the x.

Thanks!

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- Jan 13th 2011, 06:23 AMyoungb11Proving equality of two composite functions.
$\displaystyle F(x)=\sqrt{x}+1$

$\displaystyle G(x)=\sqrt{x+1}$

Prove:

$\displaystyle G(F(x))=F(g(x))$

I'm a little confused as to how I would prove x=0 and without eliminating the x.

Thanks! - Jan 13th 2011, 06:33 AMDrSteve
Let me start you off:

$\displaystyle G(F(x))=G(\sqrt{x}+1)=\sqrt{\sqrt{x}+2}$ - Jan 13th 2011, 06:40 AMyoungb11
Where did the 2 come from?

Also, I can work through it but I'm confused as how I would end up with one side with an X as the way I go through it I end up canceling the x's. Could you possibly go through it to the point where I wont have to cross out the X and end up with the same answer (0)? - Jan 13th 2011, 06:46 AMmr fantastic
- Jan 13th 2011, 07:04 AMyoungb11
Oops! I made a mistake on my original post. Th question is:

$\displaystyle F(x)=\sqrt{x}+1$

$\displaystyle G(x)=x+1$

Prove:

$\displaystyle G(F(x))=F(g(x))$

I'm a little confused as to how I would prove x=0 and without eliminating the x.

Thanks!

(The x+1 was not meant to be squarerooted) - Jan 13th 2011, 07:16 AMPlato
- Jan 13th 2011, 08:44 AMHallsofIvy
This is very confusing. you say, at first, "prove G(F(x))= F(G(x))" which implies that you want to prove it is true for all x. But then you say "prove x= 0" which implies you want to prove they are true

**only**for x= 0. Which is it? (I won't even mention using "G" and "g" interchangebly!)

I note that if x= 1, then F(1)= 1+ 1= 2 so G(F(1))= G(2)= 2+ 1= 3 while G(1)= 1+ 1= 2 so F(G(1))= F(2)= $\displaystyle \sqrt{2}+ 1\ne 3$ so you must mean "solve G(F(x))= F(G(x)) for x".

You have already been told that G(F(x))= $\displaystyle (\sqrt{x}+ 1)+ 1= \sqrt{x}+ 2$ and that F(G(x))= $\displaystyle \sqrt{x+1}+ 1$ so you need to solve

$\displaystyle \sqrt{x}+ 2= \sqrt{x+ 1}+ 1$

That involves**two**square roots so you will need to square twice. First, say, subtract 1 from both sides to isolate the $\displaystyle \sqrt{x+1}$:

$\displaystyle \sqrt{x}+ 1= \sqrt{x+ 1}$

Now square both sides to get

$\displaystyle x+ 2\sqrt{x}+ 1= x+ 1$

can you finish that? Square again get rid of this second square root.

The final equation will have**two**solutions but only one of them satisfies the original equation. That happens when you square both sides of an equation or mutliply both sides of an equation by the same thing. For example, x= 2 has only one root but $\displaystyle x^2= 4$ has two. - Jan 13th 2011, 09:48 AMmr fantastic