- 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)

- schrodinger

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- FibonacciChick666

so first, what is g composed f?

- FibonacciChick666

the function I mean,

- anonymous

g should be \[2x^2+3\]

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## More answers

- FibonacciChick666

that's what you have for f?

- anonymous

g(x)=4x-5

- FibonacciChick666

ok, so I am asking, when you plug f into g, what is the resulting function?

- FibonacciChick666

ie. \(g(f(x) ~or~ g ~o~ f\)

- 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

\[32x^2-80x+53\]

- anonymous

that was fog

- anonymous

i dont know what gof is...I need help with that I only know fog, which I just gave an answer of

- FibonacciChick666

how did you get fog, what did you do? (gof is the same process)

- FibonacciChick666

or ok, let's say let x=2, what would you do?

- FibonacciChick666

you'd put a 2 everywhere you see an x right?

- FibonacciChick666

so in the equation g(x) replace every x with f.

- FibonacciChick666

then replace every f with what f equals, it's just substitution

- anonymous

I did \[fog=f(4x-5)\]
\[fog=2(4x-5)^2+3\]
which equals \[32x^2-80x+53\]

- anonymous

that answer is for fog and now I don't know how to do gof

- UsukiDoll

for g o f you place your f(x) function into the g(x) function

- FibonacciChick666

same exact way as above

- FibonacciChick666

you did it correctly

- FibonacciChick666

its just g(f) instead of f(g)

- FibonacciChick666

do you see what I mean?

- 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

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

ah, you put the wrong function in

- anonymous

how do I put it in :(

- FibonacciChick666

you put your answer for part a in instead of doing the new problem

- FibonacciChick666

you plugged f composed with g in instead of just f

- UsukiDoll

do you see the x in g(x) = 4x- 5
replace that x in g(x) with that entire f(x) function

- FibonacciChick666

so what is f? Plug that in instead of your answer for part a, then I will double check it.

- FibonacciChick666

you did set it up correctly, just misread your paper.

- anonymous

ok so
\[g(x)=4(2x^2+3)-5\]

- FibonacciChick666

much better :)

- UsukiDoll

yes so all you have to do is distribute that 4 for 2x^2+3

- FibonacciChick666

go ahead, try and simplify

- anonymous

final answer should be
\[8x^2+7\]
right?

- UsukiDoll

we are multiplying

- FibonacciChick666

well, how do you distribute the 4?

- UsukiDoll

so what is 4 times 2x^2 and what is 4 times 3

- FibonacciChick666

so just remember, if you have \[a(c+b)=ac+ab\] You have to send the a to b and c

- anonymous

8x^2+12

- anonymous

\[8x^2+12x-5\]

- FibonacciChick666

ok now that is what happens when you distribute correctly, now, you still had that - five hanging out

- UsukiDoll

we don't need the x for the 12

- UsukiDoll

\[g(x)=8x^2+12-5\]
what is 12 - 5 ?

- anonymous

so its \[8x^2+7\]

- FibonacciChick666

yea

- FibonacciChick666

You were right to begin with, I apologize

- FibonacciChick666

I got lost in usuki's comments

- anonymous

Oh thanks...yeah I was confused that you guys said it was wrong haha

- 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

Yeah, is that the final answer?

- UsukiDoll

yeah

- FibonacciChick666

well, yes and no

- UsukiDoll

how is this no?

- FibonacciChick666

so that is the final answer for what is g composed f, but we are not just loooking for that

- FibonacciChick666

now, we need to find their domain, for both bantaray

- anonymous

we are looking for domains

- FibonacciChick666

now, what does domain mean bant?

- UsukiDoll

oh geez total miss. . .

- FibonacciChick666

I got this usuki, you can go

- anonymous

the domain for fog is \[(-\infty,\infty)\]

- FibonacciChick666

h/o I need to find what that was again haha

- FibonacciChick666

sounds right though

- FibonacciChick666

yep, that's right

- FibonacciChick666

now for g?

- FibonacciChick666

side note, you can also say All real numbers

- anonymous

Ok, I got the domains now...I figured them out really fast :) I think we are done

- FibonacciChick666

yep, what did you get for g(f)?

- anonymous

I have one more question.....Would you mind staying to help me?

- FibonacciChick666

can you pop open another question and tag me?

- FibonacciChick666

just close this one to make a new one. It's a bit of a long thread

- anonymous

yeah sure!

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