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so first, what is g composed f?
the function I mean,
g should be \[2x^2+3\]
that's what you have for f?
ok, so I am asking, when you plug f into g, what is the resulting function?
ie. \(g(f(x) ~or~ g ~o~ f\)
Just making things tidier for everyone: \[\Large f(x)=2x^2+3\]\[\Large g(x)=4x-5\]\[\Large(f\circ g)(x) = 2(4x-5)^2+3\]\[\Large(g\circ f)(x) = 4(2x^2+3)-5\] (Please correct me if I'm wrong)
that was fog
i dont know what gof is...I need help with that I only know fog, which I just gave an answer of
how did you get fog, what did you do? (gof is the same process)
or ok, let's say let x=2, what would you do?
you'd put a 2 everywhere you see an x right?
so in the equation g(x) replace every x with f.
then replace every f with what f equals, it's just substitution
I did \[fog=f(4x-5)\] \[fog=2(4x-5)^2+3\] which equals \[32x^2-80x+53\]
that answer is for fog and now I don't know how to do gof
for g o f you place your f(x) function into the g(x) function
same exact way as above
you did it correctly
its just g(f) instead of f(g)
do you see what I mean?
some people use different terminology symbols... I don't use that but I do use g o f a lot... read this from right to left... your f(x) function should be placed inside the g(x) function
I did and it keeps telling me I'm wrong. This is how I would do it... \[4(32x^2-80x+53)-5\] which would give me \[128x^2-320x+207\]
ah, you put the wrong function in
how do I put it in :(
you put your answer for part a in instead of doing the new problem
you plugged f composed with g in instead of just f
do you see the x in g(x) = 4x- 5 replace that x in g(x) with that entire f(x) function
so what is f? Plug that in instead of your answer for part a, then I will double check it.
you did set it up correctly, just misread your paper.
ok so \[g(x)=4(2x^2+3)-5\]
much better :)
yes so all you have to do is distribute that 4 for 2x^2+3
go ahead, try and simplify
final answer should be \[8x^2+7\] right?
we are multiplying
well, how do you distribute the 4?
so what is 4 times 2x^2 and what is 4 times 3
so just remember, if you have \[a(c+b)=ac+ab\] You have to send the a to b and c
ok now that is what happens when you distribute correctly, now, you still had that - five hanging out
we don't need the x for the 12
\[g(x)=8x^2+12-5\] what is 12 - 5 ?
so its \[8x^2+7\]
You were right to begin with, I apologize
I got lost in usuki's comments
Oh thanks...yeah I was confused that you guys said it was wrong haha
yea, I'm sorry about that. I read the we are multip[lying and assumed you added, I didn't notice they come out the same. But anyways, does this make sense now?
Yeah, is that the final answer?
well, yes and no
how is this no?
so that is the final answer for what is g composed f, but we are not just loooking for that
now, we need to find their domain, for both bantaray
we are looking for domains
now, what does domain mean bant?
oh geez total miss. . .
I got this usuki, you can go
the domain for fog is \[(-\infty,\infty)\]
h/o I need to find what that was again haha
sounds right though
yep, that's right
now for g?
side note, you can also say All real numbers
Ok, I got the domains now...I figured them out really fast :) I think we are done
yep, what did you get for g(f)?
I have one more question.....Would you mind staying to help me?
can you pop open another question and tag me?
just close this one to make a new one. It's a bit of a long thread