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- anonymous

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- zzr0ck3r

what is \(f(1)\)?

- anonymous

g(f(1))

- zzr0ck3r

but what is f(1)?

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## More answers

- anonymous

-4x + 7?

- zzr0ck3r

do you understand what \(f(1)\) means?

- zzr0ck3r

it means put a \(1\) in the function \(f\) where ever you see an \(x\).

- anonymous

no this is in a packet about composition of functions. Just problems to work out. I'm trying to look online.

- anonymous

not much explaination

- zzr0ck3r

but before you can understand composition of functions, you must understand a function.

- anonymous

okay

- zzr0ck3r

for the function \(f(x) = -4x+7\) we have \(f(1) = -4*1+7=3\).
Does this make sense?

- anonymous

yeah

- zzr0ck3r

ok now then, what is \(g(3)\)?

- anonymous

g(3) is we put a 3 in the function where the x is @zzr0ck3r

- anonymous

and it would look like g(3) = 2*3 - 6 I think?

- anonymous

which is 0

- Astrophysics

Hey, so the notation just implies g(f(1)) it's probably simpler to find g(f(x)) first this just means plug the function f(x) wherever there is an x in function g(x). Try that out :)

- Astrophysics

Or we can do it the way @zzr0ck3r what ever you are comfortable with!

- anonymous

i'm not sure how I would right that

- anonymous

write*

- Astrophysics

No worries, f(x) = -4x+7 and g(x) = 2x-6 so we take function f(x) and plug it in g(x) \[g(f(x)) = 2(-4x+7)-6\]

- Astrophysics

Now you can find g(f(1)) by plugging in 1 where the x is and evaluating

- Astrophysics

Does this make sense? It can be a bit confusing haha.

- anonymous

so it would be 0 then?

- anonymous

It makes sense though a bit complicated. It's something I'm gonna have to really ingrain in my head.

- anonymous

@Astrophysics how would I solve it the way @zzr0ck3r did it? I wasn't sure what to do after he left.

- Astrophysics

Yup, 0 sounds good!

- anonymous

Thanks for helping me. You're awesome :)

- Astrophysics

Ok so with zz's method you found what f(1) was which is 3 correct

- Astrophysics

Then we just take f(1) and plug that in g(x) for g(f(1)) = 2(3)-6 = 0 :)

- Astrophysics

Either way works :P

- anonymous

oh I see. Both of you guys gave me good ways. Thank you again

- Astrophysics

Yw :)

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