Please help! [find f(g(x))]

[math1325]

Hello everyone,

I am having a little problem with the following question can someone please help me out with the problem...thanks to whoever that does.


If g(x)=square root x+1 and f(x)=x^2+2, find f[g(x)].

so far my answer is this...please help.

f[g(x)]= f(square root (x+1))
= (square root (x+1))^2 +2
= 1+1+2
= 4 Is this the right answer...please help.

 

[tkhunny]

How did 'x' magically turn into '1'?

\(\displaystyle (\sqrt{x+1})^{2}\,=\,|x+1|\) ≠ (x+1)

 

[math1325]

would it be this way then....

f[g(x)]= f(square root (x+1))
= (square root (x+1))^2 +2
= (1+1)^2+2
= (square root (2))^2+2
= (square root 4)+2)
= (2)+2
= 4 as the final answer would this be right...for x magically appearing as 1 I am guessing that square root of x raise to power of 2 would equal 1 as 1/2 times 2 equals 1...therefore x is 1...I am not sure that is why I am posting this question...help if you can please.

 

[math1325]

Is my reply right or am I wrong. I think thats how you do it...my book doesn't support the example for such a problem since I have a old book, and my book store sold out of the new one...please try to help if possible...thanks a lot.

 

[tkhunny]

'x' magically turned into '1' again! Don't do that. Do you not like 'x'?

\(\displaystyle f(g(x)) = (\sqrt{x+1})^{2} + 2 = |x+1| + 2\)

It still has an 'x' in it. It's done.

 

[tkhunny]

Wow! What AM I talking about?

\(\displaystyle \sqrt{x^2}=|x|\) -- That's the right one.
\(\displaystyle (\sqrt{x})^{2} = x\) -- I can't tell you why I thought it was this one. Forget that.
\(\displaystyle \sqrt{x^2}=|x|\) -- Remember this one.

So, \(\displaystyle f(g(x)) = (\sqrt{x+1})^{2}+2 = (x+1) + 2 = x+3\)

No substitute for paying attention. I should try it.