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SephI

  • one year ago

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

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

  2. SephI
    • one year ago
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    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
    • one year ago
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    f(x) = 3? I take it you meant f(-2) indeed, f(-2) = 3 I suggest you retry g(-2)

  4. SephI
    • one year ago
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    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
    • one year ago
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    Ahh. No. \[\huge \frac{f}g(x) = \frac{f(x)}{g(x)}\\\huge \frac{f}g(-2)=\frac{f(-2)}{g(-2)}\]

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

  7. SephI
    • one year ago
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    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
    • one year ago
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    because you're not supposed to get g(f(-2)) Just g(-2)

  9. SephI
    • one year ago
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    \[\frac{ -3(-2) - 3 }{ 2(-2) + 1 }\] Is that it?

  10. terenzreignz
    • one year ago
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    Yup.

  11. SephI
    • one year ago
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    \[\frac{ 3 }{ -3 } = -1\]

  12. terenzreignz
    • one year ago
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    There you go.

  13. SephI
    • one year ago
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    ;D

  14. SephI
    • one year ago
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    Okay, and can you explain the g(f[4]) thingy?

  15. terenzreignz
    • one year ago
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    That's when you first get the result of f(4) and then plug it into g(x)

  16. SephI
    • one year ago
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    And if you had f(g[4]) You'd first get g and plug it into f?

  17. terenzreignz
    • one year ago
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    You'd first get g(4) and then plug it into f(x) , yes :)

  18. SephI
    • one year ago
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    I see. It's all about order. But sometimes it's so unclear ;-;

  19. terenzreignz
    • one year ago
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    Hopefully clearer now, if only a little bit :)

  20. SephI
    • one year ago
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    Yup! Thanks

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