anonymous
  • anonymous
Find fog and gof and their domains for the pair of functions F(x)=2x^2+3 and g(x)=4x-5 (I have fog, I cannot figure out gof)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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FibonacciChick666
  • FibonacciChick666
so first, what is g composed f?
FibonacciChick666
  • FibonacciChick666
the function I mean,
anonymous
  • anonymous
g should be \[2x^2+3\]

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FibonacciChick666
  • FibonacciChick666
that's what you have for f?
anonymous
  • anonymous
g(x)=4x-5
FibonacciChick666
  • FibonacciChick666
ok, so I am asking, when you plug f into g, what is the resulting function?
FibonacciChick666
  • FibonacciChick666
ie. \(g(f(x) ~or~ g ~o~ f\)
kc_kennylau
  • kc_kennylau
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)
anonymous
  • anonymous
\[32x^2-80x+53\]
anonymous
  • anonymous
that was fog
anonymous
  • anonymous
i dont know what gof is...I need help with that I only know fog, which I just gave an answer of
FibonacciChick666
  • FibonacciChick666
how did you get fog, what did you do? (gof is the same process)
FibonacciChick666
  • FibonacciChick666
or ok, let's say let x=2, what would you do?
FibonacciChick666
  • FibonacciChick666
you'd put a 2 everywhere you see an x right?
FibonacciChick666
  • FibonacciChick666
so in the equation g(x) replace every x with f.
FibonacciChick666
  • FibonacciChick666
then replace every f with what f equals, it's just substitution
anonymous
  • anonymous
I did \[fog=f(4x-5)\] \[fog=2(4x-5)^2+3\] which equals \[32x^2-80x+53\]
anonymous
  • anonymous
that answer is for fog and now I don't know how to do gof
UsukiDoll
  • UsukiDoll
for g o f you place your f(x) function into the g(x) function
FibonacciChick666
  • FibonacciChick666
same exact way as above
FibonacciChick666
  • FibonacciChick666
you did it correctly
FibonacciChick666
  • FibonacciChick666
its just g(f) instead of f(g)
FibonacciChick666
  • FibonacciChick666
do you see what I mean?
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
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 \[128x^2-320x+207\]
FibonacciChick666
  • FibonacciChick666
ah, you put the wrong function in
anonymous
  • anonymous
how do I put it in :(
FibonacciChick666
  • FibonacciChick666
you put your answer for part a in instead of doing the new problem
FibonacciChick666
  • FibonacciChick666
you plugged f composed with g in instead of just f
UsukiDoll
  • UsukiDoll
do you see the x in g(x) = 4x- 5 replace that x in g(x) with that entire f(x) function
FibonacciChick666
  • FibonacciChick666
so what is f? Plug that in instead of your answer for part a, then I will double check it.
FibonacciChick666
  • FibonacciChick666
you did set it up correctly, just misread your paper.
anonymous
  • anonymous
ok so \[g(x)=4(2x^2+3)-5\]
FibonacciChick666
  • FibonacciChick666
much better :)
UsukiDoll
  • UsukiDoll
yes so all you have to do is distribute that 4 for 2x^2+3
FibonacciChick666
  • FibonacciChick666
go ahead, try and simplify
anonymous
  • anonymous
final answer should be \[8x^2+7\] right?
UsukiDoll
  • UsukiDoll
we are multiplying
FibonacciChick666
  • FibonacciChick666
well, how do you distribute the 4?
UsukiDoll
  • UsukiDoll
so what is 4 times 2x^2 and what is 4 times 3
FibonacciChick666
  • FibonacciChick666
so just remember, if you have \[a(c+b)=ac+ab\] You have to send the a to b and c
anonymous
  • anonymous
8x^2+12
anonymous
  • anonymous
\[8x^2+12x-5\]
FibonacciChick666
  • FibonacciChick666
ok now that is what happens when you distribute correctly, now, you still had that - five hanging out
UsukiDoll
  • UsukiDoll
we don't need the x for the 12
UsukiDoll
  • UsukiDoll
\[g(x)=8x^2+12-5\] what is 12 - 5 ?
anonymous
  • anonymous
so its \[8x^2+7\]
FibonacciChick666
  • FibonacciChick666
yea
FibonacciChick666
  • FibonacciChick666
You were right to begin with, I apologize
FibonacciChick666
  • FibonacciChick666
I got lost in usuki's comments
anonymous
  • anonymous
Oh thanks...yeah I was confused that you guys said it was wrong haha
FibonacciChick666
  • FibonacciChick666
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?
anonymous
  • anonymous
Yeah, is that the final answer?
UsukiDoll
  • UsukiDoll
yeah
FibonacciChick666
  • FibonacciChick666
well, yes and no
UsukiDoll
  • UsukiDoll
how is this no?
FibonacciChick666
  • FibonacciChick666
so that is the final answer for what is g composed f, but we are not just loooking for that
FibonacciChick666
  • FibonacciChick666
now, we need to find their domain, for both bantaray
anonymous
  • anonymous
we are looking for domains
FibonacciChick666
  • FibonacciChick666
now, what does domain mean bant?
UsukiDoll
  • UsukiDoll
oh geez total miss. . .
FibonacciChick666
  • FibonacciChick666
I got this usuki, you can go
anonymous
  • anonymous
the domain for fog is \[(-\infty,\infty)\]
FibonacciChick666
  • FibonacciChick666
h/o I need to find what that was again haha
FibonacciChick666
  • FibonacciChick666
sounds right though
FibonacciChick666
  • FibonacciChick666
yep, that's right
FibonacciChick666
  • FibonacciChick666
now for g?
FibonacciChick666
  • FibonacciChick666
side note, you can also say All real numbers
anonymous
  • anonymous
Ok, I got the domains now...I figured them out really fast :) I think we are done
FibonacciChick666
  • FibonacciChick666
yep, what did you get for g(f)?
anonymous
  • anonymous
I have one more question.....Would you mind staying to help me?
FibonacciChick666
  • FibonacciChick666
can you pop open another question and tag me?
FibonacciChick666
  • FibonacciChick666
just close this one to make a new one. It's a bit of a long thread
anonymous
  • anonymous
yeah sure!

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