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Grazes

  • 2 years ago

Find f^-1(x) and state any restrictions on the domain of f^-1(x)

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  1. Grazes
    • 2 years ago
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    \[f(x)=\frac{ x }{ x-2 }\] \[x \neq 2 \]

  2. satellite73
    • 2 years ago
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    you need the inverse?

  3. Grazes
    • 2 years ago
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    Yes. I know I need to switch f(x) and x, but I dunno how you would simplify from there

  4. satellite73
    • 2 years ago
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    you can take \[y=\frac{x}{x-2}\] switch \(x\) and \(y\) to write \[x=\frac{y}{y-2}\] and solve for \(y\)

  5. satellite73
    • 2 years ago
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    you get \[x=\frac{y}{y-2}\]\[x(y-2)=y\] as a first step then multiply out

  6. satellite73
    • 2 years ago
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    \[xy-2x=y\] put the \(y\)'s on one side of the equal sign, the \(2x\) on the other \[xy-y=2x\]

  7. satellite73
    • 2 years ago
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    factor out the \(y\) on the left \[(x-1)y=2x\]

  8. Grazes
    • 2 years ago
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    Okay. I've got it from there

  9. Grazes
    • 2 years ago
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    I didn't think to subtract y over ^^;;

  10. satellite73
    • 2 years ago
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    and finally divide the domain will be all real numbers except \(x=1\) which is not surprising because that was the range of \(f\)

  11. satellite73
    • 2 years ago
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    answer should be \[y=\frac{2x}{x-1}\]

  12. satellite73
    • 2 years ago
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    yeah you got to get the \(y\)' all on one side good luck

  13. Grazes
    • 2 years ago
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    Thanks.

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