# Proving equality of two composite functions.

• January 13th 2011, 06:23 AM
youngb11
Proving equality of two composite functions.
$F(x)=\sqrt{x}+1$
$G(x)=\sqrt{x+1}$

Prove:
$G(F(x))=F(g(x))$

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

Thanks!
• January 13th 2011, 06:33 AM
DrSteve
Let me start you off:

$G(F(x))=G(\sqrt{x}+1)=\sqrt{\sqrt{x}+2}$
• January 13th 2011, 06:40 AM
youngb11
Quote:

Originally Posted by DrSteve
Let me start you off:

$G(F(x))=G(\sqrt{x}+1)=\sqrt{\sqrt{x}+2}$

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)?
• January 13th 2011, 06:46 AM
mr fantastic
Quote:

Originally Posted by youngb11
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)?

By the given definition, $G(F(x)) = \sqrt{F(x) + 1} = \sqrt{(\sqrt{x} + 1) + 1} = \sqrt{\sqrt{x} + 2}$.

Getting an expression for F(G(x)) and demonstrating equality with the above is left for you to do.
• January 13th 2011, 07:04 AM
youngb11
Oops! I made a mistake on my original post. Th question is:

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

Prove:
$G(F(x))=F(g(x))$

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)
• January 13th 2011, 07:16 AM
Plato
Quote:

Originally Posted by youngb11
$F(x)=\sqrt{x}+1$
$G(x)=x+1$
Prove:
$G(F(x))=F(g(x))$

$G(F(x))=(\sqrt{x}+1)+1=\sqrt{x}+2$
• January 13th 2011, 08:44 AM
HallsofIvy
Quote:

Originally Posted by youngb11
$F(x)=\sqrt{x}+1$
$G(x)=x+1$

Prove:
$G(F(x))=F(g(x))$

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

Thanks!

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)= $\sqrt{2}+ 1\ne 3$ so you must mean "solve G(F(x))= F(G(x)) for x".

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

That involves two square roots so you will need to square twice. First, say, subtract 1 from both sides to isolate the $\sqrt{x+1}$:
$\sqrt{x}+ 1= \sqrt{x+ 1}$
Now square both sides to get
$x+ 2\sqrt{x}+ 1= x+ 1$
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 $x^2= 4$ has two.
• January 13th 2011, 09:48 AM
mr fantastic
Quote:

Originally Posted by youngb11
Oops! I made a mistake on my original post. I've corrected the equations. Could anyone offer some support?

(The x+1 was not meant to be squarerooted)

It is disappointing that you have not been able to understand well enough the help given so far to attempt a solution youself.