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Thread: inverse

  1. #1
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    inverse

    dont know how to start this:

    $\displaystyle f(x) = 5 + 5 x + 4 e^x $

    find $\displaystyle f^{-1}(9) = $____
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    dont know how to start this:

    $\displaystyle f(x) = 5 + 5 x + 4 e^x $

    find $\displaystyle f^{-1}(9) = $____
    There is a theorem that says:

    Let $\displaystyle f$ be a one-to-one continuous function on an open interval $\displaystyle I$. If $\displaystyle f$ is differentiable at $\displaystyle x_0 \in I$ and if $\displaystyle f'(x_0) \neq 0$, then $\displaystyle f^{-1}$ is differentiable at $\displaystyle y_0 = f(x_0)$ and

    $\displaystyle \left( f^{-1} \right)^{ \prime} (y_0) = \frac {1}{f'(x_0)}$

    Note here, that $\displaystyle f(0) = 9$

    Let $\displaystyle x_0 = 0$ and $\displaystyle y_0 = f(0) = 9$

    Then $\displaystyle \left( f^{-1} \right)^{ \prime} (9) = \frac {1}{f'(0)}$

    Now continue
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by viet View Post
    dont know how to start this:

    $\displaystyle f(x) = 5 + 5 x + 4 e^x $

    find $\displaystyle f^{-1}(9) = $____
    $\displaystyle f(x)$ is an increasing function so if there is a solution to:

    $\displaystyle f(x)=9$

    it is unique. Now $\displaystyle f(0)=9$, so $\displaystyle f^{-1}(9) = 0$

    RonL
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    $\displaystyle f(x)$ is an increasing function so if there is a solution to:

    $\displaystyle f(x)=9$

    it is unique. Now $\displaystyle f(0)=9$, so $\displaystyle f^{-1}(9) = 0$

    RonL
    well that was simple enough. anyway, my theorem is bound to come in handy at some point in the future
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