Question

f(x)= x/(1+x)
g(x) = sin2x
Find the following (along with their domains)
a) fog
b) gof
c)fof
d) gog

Answer 1

@jim_thompson5910 help please!

Answer 2

a) fog
f(g(x)) = f(sin2x)=sin2(1/(1+x))

Answer 3

Hmm that fog looks a little mixed up :d

Answer 4

\[\large\rm f(\color{orangered}{x})=\frac{\color{orangered}{x}}{1+\color{orangered}{x}}\]Becomes:\[\large\rm f(\color{orangered}{g(x)})=\frac{\color{orangered}{\sin2x}}{1+\color{orangered}{\sin2x}}\]Ya? :o

Answer 5

What you posted: \(\large\rm sin\left(2\color{orangered}{\frac{1}{1+x}}\right)\)
This is actually the function f being plugged into g,
\[\large\rm \sin\left(2\color{orangered}{\frac{1}{1+x}}\right)=\sin\left(2\color{orangered}{f(x)}\right)=g(\color{orangered}{f(x)})\]Where,\[\large\rm \sin(2\color{orangered}{x})=g(\color{orangered}{x})\]

Answer 6

Just a second, trying to work it out on paper :)

Answer 7

haha yeah I wrote it correctly on paper but did it wrong on here xD sorry! just a moment.

Answer 8

:3

Answer 9

For part A,
do you understand how to find the domain of the composition?

Answer 10

I am only having issues with the domain for c now. The rest are all entered and correct.

Answer 11

c) x/1+2x
Domain: (-infinity, -1/2) union (-1/2, infinity)

it keeps saying the domain is wrong?

Answer 12

Oh sorry I ran off for a sec >.<

Answer 13

You're all good :)

Answer 14

Hmm, I forget how these compositions work sometimes.
I guess we need to carry this information around with us before simplifying,\[\large\rm f(x)=\frac{x}{1+x},\qquad\qquad x\ne-1\]So when we get to this composition we have,\[\large\rm f(f(x))=\frac{x}{1+2x},\qquad\qquad x\ne-1,-\frac{1}{2}\]Try that maybe? :o

Answer 15

aaaah.
so in interval notation it would be: (-infinity, -1) union (-1,-1/2)union(-1/2, infinity)

Answer 16

Mmmmm yah I think so :)