Question

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)

Answer 1

so first, what is g composed f?

Answer 2

the function I mean,

Answer 3

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

Answer 4

that's what you have for f?

Answer 5

g(x)=4x-5

Answer 6

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

Answer 7

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

Answer 8

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)

Answer 9

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

Answer 10

that was fog

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

same exact way as above

Answer 21

you did it correctly

Answer 22

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

Answer 23

do you see what I mean?

Answer 24

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

Answer 25

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
\[128x^2-320x+207\]

Answer 26

ah, you put the wrong function in

Answer 27

how do I put it in :(

Answer 28

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

Answer 29

you plugged f composed with g in instead of just f

Answer 30

do you see the x in g(x) = 4x- 5

replace that x in g(x) with that entire f(x) function

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

much better :)

Answer 35

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

Answer 36

go ahead, try and simplify

Answer 37

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

Answer 38

we are multiplying

Answer 39

well, how do you distribute the 4?

Answer 40

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

Answer 41

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

Answer 42

8x^2+12

Answer 43

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

Answer 44

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

Answer 45

we don't need the x for the 12

Answer 46

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

Answer 47

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

Answer 48

yea

Answer 49

You were right to begin with, I apologize

Answer 50

I got lost in usuki's comments

Answer 51

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

Answer 52

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?

Answer 53

Yeah, is that the final answer?

Answer 54

yeah

Answer 55

well, yes and no

Answer 56

how is this no?

Answer 57

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

Answer 58

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

Answer 59

we are looking for domains

Answer 60

now, what does domain mean bant?

Answer 61

oh geez total miss. . .

Answer 62

I got this usuki, you can go

Answer 63

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

Answer 64

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

Answer 65

sounds right though

Answer 66

yep, that's right

Answer 67

now for g?

Answer 68

side note, you can also say All real numbers

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

can you pop open another question and tag me?

Answer 73

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

Answer 74

yeah sure!