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hpfan101

  • one year ago

Find a formula for the inverse of the function. f(x)=(4x-1)/(2x+3) y=ln(x+3) y=x^2 - x y=(1+e^-x)/(1+e^-x)

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  1. freckles
    • one year ago
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    if the function is one-to-one, you just solve the equation for x

  2. freckles
    • one year ago
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    \[y=\frac{ax+b}{cx+d} \\ \text{ multiply both sides by } cx+d \\ (cx+d)y=ax+b \\ \text{ distribute on the left } \\ cx y +dy=ax+b \\ \text{ now remember you want to solve for } x \\ \text{ get all of your terms with } x \text{ on one side } \\ \\ \text{ so subtract } ax \text{ and } dy \text{ on both sides } \\ cxy-ax=b-dy \\ \text{ now this allows you to do the following so you can solve for } x \\ x(cy-a)=b-dy \\ \text{ now divide both sides by } (cy-a)\]

  3. hpfan101
    • one year ago
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    Oh, ok. Thank you! :)

  4. freckles
    • one year ago
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    after solving for x,interchange x any y

  5. hpfan101
    • one year ago
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    y=(1+e^-x)/(1+e^-x) How about the very last one? The negative exponent is confusing me.

  6. freckles
    • one year ago
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    if the negative exponent confused you can multiply top and bottom by e^x

  7. freckles
    • one year ago
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    \[y=\frac{1+e^{-x}}{1+e^{-x}} \cdot \frac{e^{x}}{e^{x}} \\ y=\frac{e^{x}+1}{e^{x}+1}\]

  8. freckles
    • one year ago
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    wait a minute

  9. freckles
    • one year ago
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    lol (1+exp(-x))/(1+exp(-x)) is 1

  10. freckles
    • one year ago
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    is there a type-o?

  11. freckles
    • one year ago
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    \[y=\frac{1+e^{-x}}{1+e^{-x}}=1 \]

  12. freckles
    • one year ago
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    and f(x)=1 is not one-to-one

  13. hpfan101
    • one year ago
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    I think the typo is that the numerator is (1-e^-x) while the denominator is (1+e^-x). But would that make that much of a difference?

  14. hpfan101
    • one year ago
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    Oh wait, it would. You can't simplify it to be 1.

  15. freckles
    • one year ago
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    yes! now the top is different from the bottom the function is not 1 anymore now verify the function is one-to-one and if it is solve for e^x and then solve for x and if is not you are done because no inverse function exist

  16. freckles
    • one year ago
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    I said solve for e^x because you can still multiply top and bottom by e^x giving you \[f(x)=\frac{e^{x}-1}{e^{x}+1}\]

  17. hpfan101
    • one year ago
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    Ah I see how to solve it now. Thanks again!

  18. freckles
    • one year ago
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    are you looking for inverse relation or function?

  19. hpfan101
    • one year ago
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    Inverse function

  20. freckles
    • one year ago
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    so did you verify f(x)=(1-exp(-x))/(1+exp(-x)) is one-to-one? you can use a graphing calculator and see if it passes the horizontal line test?

  21. hpfan101
    • one year ago
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    Oh ok. So I plugged it into the graphing calculator and the function doesn't pass he horizontal line test.

  22. hpfan101
    • one year ago
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    the*

  23. hpfan101
    • one year ago
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    Actually, it does. Nevermind, it looked like it didn't at first.

  24. freckles
    • one year ago
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    ok so that means you have to actually solve that equation above for e^x then x

  25. hpfan101
    • one year ago
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    Alright, will do.

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