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iop360

  • 3 years ago

isolate y for the following equation:

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  1. iop360
    • 3 years ago
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    \[x= -14e ^{2-y} + y^3 + 7y + 6\]

  2. ByteMe
    • 3 years ago
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    i don't think there is an elementary method to do that.

  3. iop360
    • 3 years ago
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    hmm the original equation asks to find the inverse function of \[-14e ^{2-x} + x^3 + 7x + 6 \]

  4. iop360
    • 3 years ago
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    which means swapping y for x and solving for y

  5. iop360
    • 3 years ago
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    and then find the equation of the line perpendicular to this inverse function at x = 14

  6. ByteMe
    • 3 years ago
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    correct... but isolating the y after you reversed it is very difficult. i guess all you could do here is give the function implicitly.

  7. ByteMe
    • 3 years ago
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    can you give the original problem? looks like you need to take derivative implicitly here.

  8. iop360
    • 3 years ago
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    differentiate both sides of the equation, find y' maybe

  9. iop360
    • 3 years ago
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    ok

  10. iop360
    • 3 years ago
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    let \[f(x) = -14e ^{2-x} + x^3 + 7x + 6\] find an equation of the normal line to the graph of f-1(x) at the point on the graph where x=14

  11. iop360
    • 3 years ago
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    normal line btw, means the line w/ a negative reciprocal

  12. iop360
    • 3 years ago
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    aka perpendicular. And the answer to this is y = -33x+464

  13. ByteMe
    • 3 years ago
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    do you know how the derivatives (slopes) of inverse functions are related?

  14. iop360
    • 3 years ago
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    maybe could you state it?

  15. iop360
    • 3 years ago
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    i get \[\frac{ dy }{ dx } = \frac{ 1 }{ 14e ^{2-y} + 3y^2 + 7 }\]

  16. iop360
    • 3 years ago
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    my prof gave a hint saying we had to find a value for x that would make the \[e ^{2-x} \] part equal to 0. dont know if that helps tho

  17. hartnn
    • 3 years ago
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    no, need to find the inverse function, u need to find y when x=14 lets say y=2 so u get \(f^{-1}(14)=2\) then (14,2) is on the graph of x = f-1(y). The slope of that function at that point is what you are looking for.

  18. iop360
    • 3 years ago
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    how do we find\[f ^{-1}(14)\]

  19. ByteMe
    • 3 years ago
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    since f(2)=14, then f-1(14)=2

  20. iop360
    • 3 years ago
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    oh right

  21. iop360
    • 3 years ago
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    wasnt 2 just used as a hypothetical

  22. hartnn
    • 3 years ago
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    yeah, thats a challenge \(14 = -14e ^{2-x} + x^3 + 7x + 6\) solve for x to get x=2 is not easy

  23. hartnn
    • 3 years ago
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    i found it by trial and error method

  24. iop360
    • 3 years ago
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    oh ok

  25. hartnn
    • 3 years ago
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    the 1st thing i thought is to eliminate e, by putting x=2 and voila!

  26. iop360
    • 3 years ago
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    oh i see

  27. hartnn
    • 3 years ago
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    so now u have a point, the derivative,can u find the slope of line....and then the equation?

  28. iop360
    • 3 years ago
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    yes that makes it way easier

  29. iop360
    • 3 years ago
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    thanks to both of you

  30. hartnn
    • 3 years ago
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    *point and derivative of inverse function....

  31. hartnn
    • 3 years ago
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    welcome ^_^

  32. ByteMe
    • 3 years ago
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    yw... also remember you're not looking for the tangent but the normal line

  33. hartnn
    • 3 years ago
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    i got this because i have seen a similar question here, http://www.physicsforums.com/showthread.php?t=397539

  34. iop360
    • 3 years ago
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    i see

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