## SephI 3 years ago Can someone PLEASE explain how to solve this? The fraction at the end completely throws me off!! Pic related: http://prntscr.com/12xgei

1. terenzreignz

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

2. SephI

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.

3. terenzreignz

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

4. SephI

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.

5. terenzreignz

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

6. terenzreignz

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

7. SephI

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?

8. terenzreignz

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

9. SephI

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

10. terenzreignz

Yup.

11. SephI

$\frac{ 3 }{ -3 } = -1$

12. terenzreignz

There you go.

13. SephI

;D

14. SephI

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

15. terenzreignz

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

16. SephI

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

17. terenzreignz

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

18. SephI

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

19. terenzreignz

Hopefully clearer now, if only a little bit :)

20. SephI

Yup! Thanks