Question

Please Help me

Given that f(x)=1/x and g(x)=8 x - 4, calculate
g o f(x)
f o f(x)
g o g(x)

when i solved g o f(x) I ended up with 8(1/x)-4 and that came to 8x-4...which i got wrong
For the other twoo I wasn't sure where to even start. won't they just equal eachother? wouldn't f o f(x) just be 1/x=1/x ? I got that wrong too. Help please.

Answer 1

\[ f(x)=\frac1x,\qquad g(x)=8 x - 4\]

\[g\circ f(x)=g(f(x))=g\left(\frac1x\right)=8\left(\frac1x\right)-4=\frac8x-4\]
\[f\circ f(x)=f\left(f(x)\right)=f\left(\frac1x\right)=\frac1{\frac1x}=x\]

Answer 2

Thank you so much. I see how you did the last one, I think my brain is just trying to make sense of it

Answer 3

I though that it would be 1/x=1/x for some reason

Answer 4

what do you get for \(g\circ g(x)\)

Answer 5

I did the same thing and got it wrong. I just did 8x-4=8x-4 ...but I'm thinking the set up may be more like 8(8x-4)-4 ...is that right? then my answer would be (64x-32)-4 which would be 64x-28? I may have just over complicated that

Answer 6

for the g o f if the x is in the denominator, then wouldn't your domain be all real numbers except -8?

Answer 7

probably 0 would make more sense

Answer 8

\[g\circ g(x)=g(8x-4)=8(8x-4)-4=64x-32-4\] is right so far but -32-4 is not -28, try that bit again

Answer 9

so 64x-36

Answer 10

\[\checkmark\]

Answer 11

Thank you so much for taking the time to explain this to me. I was having a serious meltdown trying to understand all this stuff for an exam in 3 weeks. Your are amazing! Thank you so much!!

Answer 12

now should we look at the domains?

Answer 13

oh gawd...i feel stressed already. yes we should

Answer 14

so for g o f i think the domain would be all real numbers except 0, because it can't zero out right? so it would be (-infinity, 0] ?

Answer 15

all real numbers except 0, is right but you havent got the interval notation right

(-infinity, 0] is only half of the domain

Answer 16

for f of f I think all real numbers except zero...so (-infinity, 0] or would it be (-infinity, 0) since you don't want it to be 0?

Answer 17

i don't understand, what do you mean by half of the domain?

Answer 18

x can be positive too

Answer 19

so then...umm. (-infinity, 0], [0, infinity)

Answer 20

or would i use the U since fit can't be zero?

Answer 21

the domain of

f(x)=1/x

is

(-∞,0)U(0,∞)

square brackets mean the boundary term Is included, ( and it should't be in this case )

Answer 22

that makes sense. so i only include the bracket sign when the actual number is included? but isn't the domain always "what can't equal zero" so then wouldn't everything under the domain really be enclosed in ( )

Answer 23

what i just wrote confused me. so i'll move on to the domain of g o g now

Answer 24

so it would be: (-infinity, infinity) ...right?

Answer 25

Well you have got the right domain for g o g , yes (all real numbers)

Answer 26

the domain for f o f is a little trickier

Answer 27

whew. i'm glad something made sense. Thank you so much for your help.

Answer 28

wouldn't it just be ..all real numbers?

Answer 29

I want to say x for some reason

Answer 30

\[f\circ f(x)=\frac1{\frac1x}=x\]


x cannot be zero

Answer 31

I don't understand how you came up with that.

Answer 32

I know it can't be "x" bc that isn't an actual number, it seems like -(x/1) would make more sense

Answer 33

this page might explain it better than i can , http://www.regentsprep.org/Regents/math/algtrig/ATP7/domaincomposite.htm

Answer 34

Thank you! I'll be printing that out about now. well it's past midnight here and I have to drag myself to work in the morning so I need to get off, but thank you so much for the help and for the link to that website. I really appreciate it.

Answer 35

Khan academy is another good resource when you have time

https://www.khanacademy.org/math/algebra/algebra-functions

good night and good luck

Answer 36

Thank you very much!