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Adi3

  • one year ago

Will medal Please help. g(x) = (x)/x-5 (i) find g^-1(x) (ii) find g(6) and g^-1(1)

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  1. Adi3
    • one year ago
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    @butterflydreamer

  2. Adi3
    • one year ago
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    @sammixboo @ganeshie8

  3. Adi3
    • one year ago
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    @fishejac000

  4. anonymous
    • one year ago
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    yes

  5. butterflydreamer
    • one year ago
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    oh gosh xD I always get confused when i do inverse functions. Firstly, let's start off with our function: \[g(x) = \frac{ x }{ x - 5 }\] Rename g(x) as "y" \[y = \frac{ x }{ x - 5}\] Multiply both sides by (x - 5) \[y (x - 5) = x\] Expand . \[xy - 5y = x\] move all the x's to one side. \[xy - x = 5y\] Factorise out the "x". \[x ( y - 1) = 5y\] Make "x" the subject. \[x = \frac{ 5y }{ y - 1}\] Now to find the inverse, simply swap the x's with y's and vice versa :) Then rename the function as it's inverse \[g ^{-1} (x)\]

  6. Adi3
    • one year ago
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    so i get \[\left(\begin{matrix}5x \\ x-1\end{matrix}\right)\]

  7. butterflydreamer
    • one year ago
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    er.. if that's a fraction then yes LOL. \[g^{-1} (x) = \frac{ 5x }{ x-1 } \]

  8. Adi3
    • one year ago
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    yes, what about the (ii) one

  9. butterflydreamer
    • one year ago
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    For part (II). We know that \[g(x) = \frac{ x }{ x - 5 }\] so to find g(6) sub x = 6 and we get: \[g(6) = \frac{ 6 }{ 6 - 5 } = ?\] Do the same for g^-1(x) but sub x = 1 \[g^{-1} (x) = \frac{ 5x }{ x-1} \rightarrow g^{-1} (1) = \frac{ 5(1) }{ 1 - 1 } = ?\]

  10. Adi3
    • one year ago
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    5

  11. Adi3
    • one year ago
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    how about the g^-1(1)

  12. anonymous
    • one year ago
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    5 give me my medal

  13. butterflydreamer
    • one year ago
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    ?? take a look at what i typed in my prev. post :) i showed you both g(6) and g^-1 (1)

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