F of G problem: f(x) = 1 + x^2, g(x) = sqrt[x - 1]

[kpx001]

I forgot the rules to apply when multiplying w/ square roots.

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

f o g = 1 + (sqroot[x-2]) --> do i square everything to get 1 + x - 2 --> x-1 ??

g(f(x)) = [sqroot(1+x^2)-1] then square it all to get 1+x^2 - 1 ---> x^2?

(f o g) (2) = same thing but multiply final to get 2(x-1) --> 2x-2 ??

sorry but im rusty on my math and have no books or review

 

[tkhunny]

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

It's important to note that sqrt(x-1) is POSITIVE. You tell me why.

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

You do g(f(x))

 

[kpx001]

g(f(x)) would be

(sqroot[1+x^2]-1)
then square it to get

1+x^2-1
and final answer is x^2 ??

and for (f o g)(2) is 2x ?

 

[skeeter]

no.

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

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

so ... \(\displaystyle \L g[f(2)] = |2| = 2\)

 

[tkhunny]

(sqroot[1+x^2]-1)

Good. I would have preferred writing the whole thing. Let the notation help you. g(f(x)) = (sqroot[1+x^2]-1)

then square it to get

Why did you do that? If someone owe you $10, can you just square it an make your debtor pay you $100? (Ignoring the square units, of course.) You can't just go around squaring things for fun. Never do that again. (You still can see the friendly smile on my face, right?)

1+x^2-1
and final answer is x^2 ??

This step is quite good, except for the fact that it started in the wrong place (see previous step). Also, when you do your algebra correctly, why do you doubt?