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Math Help - derivative of inverse

  1. #1
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    derivative of inverse

    if f(x)=x+\frac{1}{x} find (f^{-1})'(\frac{17}{4})

    I know that i have to use the formula

    [f^{-1}]'(a)=\frac{1}{f'[f^{-1}(a)]}

    but i am unable to find the inverse.
    Last edited by yoman360; March 19th 2010 at 08:34 PM.
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  2. #2
    Super Member Random Variable's Avatar
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     f(x) = x + \frac{1}{x}

     f'(x) = 1 - \frac{1}{x^{2}}

    so  (f^{-1})'(x+ \frac{1}{x}) = \frac{1}{1 - \frac{1}{x^{2}}}

    notice that  f(4) = 4 + \frac{1}{4} = \frac{17}{4}

    so  (f^{-1})'\Big(\frac{17}{4}\Big) = \frac{1}{1- \frac{1}{16}} = \frac{16}{15}
    Last edited by Random Variable; March 19th 2010 at 08:00 PM.
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  3. #3
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    Hello, yoman360!

    This has nothing to do with derivatives . . .


    If f(x)\:=\:x+\frac{1}{x}. find f^{-1}\left(\frac{17}{4}\right)
    By inspection, we see that: . \begin{array}{ccc}f(4) &=& \dfrac{17}{4} \\ \\[-3mm] f\left(\dfrac{1}{4}\right) &=& \dfrac{17}{4} \end{array}

    Therefore: . f^{-1}\left(\frac{17}{4}\right) \;=\;4,\;\frac{1}{4}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If we must find the inverse function first . . .

    We have: . y \;=\;x + \frac{1}{x}

    Switch varoables: . x \;=\;y + \frac{1}{y}

    Solve for y\!:\;\;xy \;=\;y^2 + 1 \quad\Rightarrow\quad y^2 - xy + 1 \:=\:0

    Quadratic Formula: . y \:=\:\frac{x \pm\sqrt{x^2-4}}{2}

    Hence: . f^{-1}(x) \;=\;\frac{x\pm\sqrt{x^2-4}}{2}


    Therefore: . f^{-1}\left(\frac{17}{4}\right) \;=\;\frac{\frac{17}{4} \pm\sqrt{\left(\frac{17}{4}\right)^2-4}}{2} \;=\;\frac{\frac{17}{4}\pm\sqrt{\frac{225}{16}}}{2  }

    . . . . . =\;\frac{\frac{17}{4} \pm\frac{15}{4}}{2} \;=\; \begin{Bmatrix}<br />
\dfrac{\frac{17}{4} + \frac{15}{4}}{2} &=& \dfrac{\frac{32}{4}}{2} &=& \dfrac{8}{2} &=& 4 \\ \\[-3mm]<br />
\dfrac{\frac{17}{4}-\frac{15}{4}}{2} &=& \dfrac{\frac{2}{4}}{2} &=& \dfrac{\frac{1}{2}}{2} &=& \dfrac{1}{4} \end{Bmatrix}

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  4. #4
    Super Member Random Variable's Avatar
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    Wait. Are you looking for the derivative of the inverse at 17/4 or just the inverse at 17/4?
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  5. #5
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    Quote Originally Posted by Random Variable View Post
    Wait. Are you looking for the derivative of the inverse at 17/4 or just the inverse at 17/4?
    derivative. I forgot to put the ' in the question
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by yoman360 View Post
    derivative. I forgot to put the ' in the question
    Then it should be what I posted above.
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  7. #7
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    It's actually kind of a bullshit question because the function is not one-to-one, and therefore does not have an inverse. Was there a domain restriction given to make it a 1-1 function?
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by drumist View Post
    It's actually kind of a bullshit question because the function is not one-to-one, and therefore does not have an inverse. Was there a domain restriction given to make it a 1-1 function?
    Yeah, that's true.


    Depending on the domain restrictions,

     (f^{-1})'(\frac{17}{4}) = \frac{1}{1 + \frac{1}{1/16}} = \frac{1}{17}

    or  (f^{-1})'(\frac{17}{4}) = \frac{16}{15}
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  9. #9
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    Quote Originally Posted by drumist View Post
    It's actually kind of a bullshit question because the function is not one-to-one, and therefore does not have an inverse. Was there a domain restriction given to make it a 1-1 function?
    Good point.
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  10. #10
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    Quote Originally Posted by Random Variable View Post
     f(x) = x + \frac{1}{x}

     f'(x) = 1 - \frac{1}{x^{2}}

    so  (f^{-1})'(x+ \frac{1}{x}) = \frac{1}{1 - \frac{1}{x^{2}}}

    notice that  f(4) = 4 + \frac{1}{4} = \frac{17}{4}

    so  (f^{-1})'\Big(\frac{17}{4}\Big) = \frac{1}{1- \frac{1}{16}} = \frac{16}{15}
    Thanks your answer is correct and has helped me to understand this. Thank you
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