Question

Medal for UR help:

Answer 1

here we have to write the function g(f(x)). What is the function g(f(x))?

Answer 2

f is the f(x) on and g is the (x) one

Answer 3

I think it wonts you to combine them

Answer 4

yes!

Answer 5

hint:
\[\Large g\left( {f\left( x \right)} \right) = \sqrt {f\left( x \right) + 1} = ...?\]

Answer 6

you have to replace f(x) with its definition, or formula

Answer 7

crap i forgot

Answer 8

ok I got another fraction

Answer 9

hint:
\[\Large g\left( {f\left( x \right)} \right) = \sqrt {f\left( x \right) + 1} = \sqrt {\frac{1}{{{x^2} - 1}} + 1} = ...?\]

Answer 10

(x+2) and (x-1)

Answer 11

we have to simplify that expression, as below:
\[\Large \begin{gathered}
g\left( {f\left( x \right)} \right) = \sqrt {f\left( x \right) + 1} = \sqrt {\frac{1}{{{x^2} - 1}} + 1} = \hfill \\
= \sqrt {\frac{{{x^2}}}{{{x^2} - 1}}} \hfill \\
\end{gathered} \]
Now the last radical exists, if the subsequent inequality is checked:
\[\Large \frac{{{x^2}}}{{{x^2} - 1}} \geqslant 0\]

Answer 12

so you have to solve this inequality:
\[\Large \frac{{{x^2}}}{{{x^2} - 1}} \geqslant 0\]

Answer 13

do you know how to solve that inequality?

Answer 14

since the numerator is always positive and it is zero at x=0, then we have to solve this equivalent inequality:
\[\Large {x^2} - 1 > 0\]

Answer 15

i can so it its :

Answer 16

ok so we can write this:
domain of g(f(x)) is the intersection between these two sets:
real line and (-infinity, -1) union (1,+infinity)
so we have:

so the requested domain is:
\[\left( { - \infty , - 1} \right) \cup \left( {1, + \infty } \right)\]