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gjhfdfg

Finding inverses to the one-to-one functions,

  • one year ago
  • one year ago

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  1. gjhfdfg
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    This is what I'm working on, |dw:1361136194386:dw|

    • one year ago
  2. gjhfdfg
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    I thought it was 7/2x-5 but I was wrong

    • one year ago
  3. poopsiedoodle
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    -7/-2x-5? just a guess. I'm not sure.

    • one year ago
  4. gjhfdfg
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    There arent any -7's in it

    • one year ago
  5. poopsiedoodle
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    well I know, but -7 is the opposite of 7. There aren't any -5s in it either. That looks like a +5 up there.

    • one year ago
  6. gjhfdfg
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    Hmm

    • one year ago
  7. gjhfdfg
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    I don't think it necessarily wants the opposite of everything

    • one year ago
  8. zenai
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    switch x and y and solve for y for the inverse of a function. \[x = \frac{ 2y+5 }{ 7}\]

    • one year ago
  9. Meepi
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    \[y = \frac{2x + 5}{7}\] To get the inverse, solve for x, then swap x and y \[y = \frac{2x + 5}{7}\] \[7y - 5 = 2x\] \[x = \frac{7y - 5}{2}\] So the inverse function is \[f^{-1}(x) = \frac{7x - 5}{2}\]

    • one year ago
  10. zenai
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    ^ he's right :P

    • one year ago
  11. gjhfdfg
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    Got it thanks. ^^

    • one year ago
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