Can someone PLEASE explain how to solve this? The fraction at the end completely throws me off!!
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- anonymous

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- terenzreignz

Okay...
So, essentially
\[\huge \frac{f}g(x) = \frac{f(x)}{g(x)}\] is the key to understanding this...

- anonymous

I deducted it and f(x) = 3
and g(x) = 7
But, my answer isn't 7... My answers are:
3
-3
-1
1
I tried dividing 3 by 7 and 7 by 3, but that didn't work.

- terenzreignz

f(x) = 3?
I take it you meant
f(-2)
indeed,
f(-2) = 3
I suggest you retry g(-2)

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## More answers

- anonymous

I thought:
\[\frac{ f }{ g }(-2) = \frac{ f(-2) }{ g }\]
Then I'd find f(-2)
Then I'd plug in -2 to g(x) and get 7.

- terenzreignz

Ahh. No.
\[\huge \frac{f}g(x) = \frac{f(x)}{g(x)}\\\huge \frac{f}g(-2)=\frac{f(-2)}{g(-2)}\]

- terenzreignz

It does bug me when they use shorthand notation resulting into unclear stuff... but yeah, that's Maths for you :P

- anonymous

Here is what I did:
f(x) = -3x - 3
f(x) = -3(-2) - 3
f(x) = 6 - 3
f(x) = 3
Plug it into g
g(x) = 2(3) + 1
g(x) = 7
But the answer isn't seven. Am I doing it wrong?

- terenzreignz

because you're not supposed to get g(f(-2))
Just g(-2)

- anonymous

\[\frac{ -3(-2) - 3 }{ 2(-2) + 1 }\]
Is that it?

- terenzreignz

Yup.

- anonymous

\[\frac{ 3 }{ -3 } = -1\]

- terenzreignz

There you go.

- anonymous

;D

- anonymous

Okay, and can you explain the g(f[4]) thingy?

- terenzreignz

That's when you first get the result of f(4) and then plug it into g(x)

- anonymous

And if you had f(g[4]) You'd first get g and plug it into f?

- terenzreignz

You'd first get g(4) and then plug it into f(x) , yes :)

- anonymous

I see. It's all about order. But sometimes it's so unclear ;-;

- terenzreignz

Hopefully clearer now, if only a little bit :)

- anonymous

Yup! Thanks

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