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Thread: Inverse function

  1. #1
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    Inverse function

    f(x)=x^3 + 2x -1

    How do I find the inverse function?
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  2. #2
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    Quote Originally Posted by Stuck Man View Post
    f(x)=x^3 + 2x -1

    How do I find the inverse function?
    not the easiest thing to find ... is there a specific reason for finding the inverse? in other words, is this need for an inverse function tied to another question?
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  3. #3
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    It ask s for the value of the inverse function at 2. It then asks for the value of the derivative of the inverse function at 2.
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    Quote Originally Posted by Stuck Man View Post
    It ask s for the value of the inverse function at 2. It then asks for the value of the derivative of the inverse function at 2.
    that's what I thought ... you do not need the inverse function to do this problem.

    f(x) = x^3 + 2x - 1

    note that f(1) = 2 , so if g(x) is the inverse of f(x), then g(2) = 1


    further, if two functions, f(x) and g(x), are inverses, then f[g(x)] = x

    \frac{d}{dx} \left(f[g(x)] = x\right)<br />

    f'[g(x)] \cdot g'(x) = 1

    g'(x) = \frac{1}{f'[g(x)]}<br />

    g'(2) = \frac{1}{f'[g(2)]}

    g'(2) = \frac{1}{f'[g(2)]}

    g'(2) = \frac{1}{f'(1)}

    find f'(1) and finish it.
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    I've not done any calculus like that so I don't understand it but I might someday.
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  6. #6
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    What do you mean by "any calculus like that"? If you are asked this question, surely you have worked with derivatives and the chain rule? If g is the inverse function to f, then f(g(x))= x so, by the chain rule, f'(g(x))g'(x)= 1. g(2)= 1 because f(x)= x^3+ 2x- 1= 2 has 1 as an obvious root. f'(g(2))= f'(1) which is 3(1)^2+ 2= 5. So 5g'(2)= 1, g'(2)= 1/5.
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