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Math Help - Inverse Derivative

  1. #1
    VkL
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    Inverse Derivative

    For f(x) = 2x^5 + 3x^3 + x

    Fi nd (f^-1)' (6)


    How do you do this?
    The formula is d/dx (f^-1(x)) = 1/ (f'(f^-1(x))

    How do I set it up?

    Thanks.
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  2. #2
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    As soon as you're given f(x), you can find its derivative f'(x)=10x^{4}+9x^{2}+1. You also need to find the inverse of f evaluated at 6: f^{-1}(6)=1 by inspection.

    Now you just plug that information into your equation \left.\begin{matrix}\frac{d}{dx}f^{-1}(x) \end{matrix}\right|_{x=6}=\frac{1}{f'(f^{-1}(6))} to get the answer.
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  3. #3
    VkL
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    Can you please show how you got f^-1(6) = 1
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  4. #4
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    To find f^{-1}(6), you need to find a value for x that makes y equal to 6. That is, you need to find the solution for 2x^5 + 3x^3 + 1 = 6.

    Notice that the coefficients of f(x) add up to 6. Knowing this, all you have to do is plug in x=1 to make the x's "disappear":

    2(1)^5 + 3(1)^3 + 1 = 2 + 3 + 1 = 6.

    A general way to solve this type of problem is to use synthetic division to factor the polynomial:
    <br />
y=2x^5 + 3x^3 + 1=6<br />
    <br />
2x^5 + 3x^3 - 5 = 0<br />
    <br />
(x-1)(2x^4+2x^3+5x^2+5x+5)=0<br />
    <br />
x=1<br />
.
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