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

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

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

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

Yup.

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

There you go.

;D

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

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

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

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

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

Hopefully clearer now, if only a little bit :)

Yup! Thanks