Question

HELP!! @waterineyes

Answer 1

what do you not know how to do?

Answer 2

haha

\[\sqrt[3]{4}=_{3}^{5}\]

Answer 3

Will go very typical..

Answer 4

\[f(x) = \frac{1-x}{1+x} \\ g(x) = \frac{x}{1-x}\]

Answer 5

Now, see, in f(x), if I change x to y, then it becomes:
\[f(y) = \frac{1-y}{1+y}\]
Getting this?

Answer 6

Instead of y, it I would change it with g(x), remember replace x with g(x), then it will look like:
\[f(g(x)) = \frac{1-g(x)}{1+g(x)}\]

Answer 7

you know value of g(x), substitute in it..

Answer 8

\[f(g(x)) = \frac{1 - \frac{x}{1-x}}{1 + \frac{x}{1-x}}\]

Answer 9

Got this step??

Answer 10

yess

Answer 11

Make denominators equal and solve it now..

Answer 12

\[f(g(x)) = \frac{1 - \frac{x}{1-x}}{1 + \frac{x}{1-x}} = \frac{1-2x}{1} = 1-2x\]

Answer 13

There is no restriction on \(x\), so :
\[Domain : \mathbb{R}\]

Answer 14

domain is all real numbers?

Answer 15

Domain is: All Real Numbers..

Answer 16

Sorry, I think that domain is:
\[\mathbb{R} -\left\{ 1 \right\}\]
because in x=1, g(x) not exists

Answer 17

@Michele_Laino has said it rightly.. :)
Sorry for that mistake.. @wade123

Answer 18

Thanks for correcting me.. :)

Answer 19

@waterineyes do not worry, we are only discussing between friends!

Answer 20

@wade thanks!

Answer 21

@wade123 yes I think!