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Math Help - finding inverse

  1. #1
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    finding inverse

    f(x)=x+3x+6

    i need to find f^-1'(2)

    do i need to find the inverse of f(x) before i derive it. or do i derive it first, then find the inverse, and then plug 2 in for x?
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  2. #2
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    Quote Originally Posted by johnnyc418 View Post
    f(x)=x+3x+6

    i need to find f^-1'(2)

    do i need to find the inverse of f(x) before i derive it. or do i derive it first, then find the inverse, and then plug 2 in for x?

    I suppose you already know that f^{-1}{}'=\frac{1}{f'(x)} , when the derivative in the denominator is expressed in the same variable as the one of f^{-1} , of course.

    So in this case f^{-1}{}'(y)=\frac{1}{3x^2+3} , when x is expressed as function of y, and we don't have the faintest idea what is this...but we don't need to have: just check that as f(x)=2\Longleftrightarrow\,x=-1 , then f^{-1}(2)=-1\Longrightarrow\,f^{-1}{}'(2)=\frac{1}{3(-1)^2+3}=\frac{1}{6}

    Tonio
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  3. #3
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    Quote Originally Posted by johnnyc418 View Post
    f(x)=x+3x+6

    i need to find f^-1'(2)

    do i need to find the inverse of f(x) before i derive it. or do i derive it first, then find the inverse, and then plug 2 in for x?
    f[f^{-1}(x)] = x

    f'[f^{-1}(x)] \cdot {f^{-1}}'(x) = 1

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


    now, use the fact that f(-1) = 2 and f^{-1}(2) = -1 to find {f^{-1}}'(2)
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  4. #4
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    Quote Originally Posted by skeeter View Post
    f[f^{-1}(x)] = x

    f'[f^{-1}(x)] \cdot {f^{-1}}'(x) = 1

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


    now, use the fact that f(-1) = 2 and f^{-1}(2) = -1 to find {f^{-1}}'(2)

    the second post confused me. is f^-1'(2)=1/6. i didnt quite understand the last part
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  5. #5
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    {f^{-1}}'(2) = \frac{1}{f'[f^{-1}(2)]}

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

    since f(x) = x^3+3x+6

    f'(x) = 3x^2 + 3

    f'(-1) = 3(-1)^2 + 3 = 6

    so ...

    {f^{-1}}'(2) = \frac{1}{f'(-1)} = \frac{1}{6}


    unconfused now?
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  6. #6
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    yes thank you
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