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Math Help - Problem with the inverse function

  1. #1
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    Problem with the inverse function

    Given that g is the inverse function of f, where:

    f(x) = \int^x_0 \frac {1}{\sqrt{t^3+1}} dt

    for f: [0,\infty) \rightarrow \mathbb{R}


    I have to show that:

    g''(x) = \frac{3}{2} (g(x))^2


    At first I thought that all I had to do was to use the relation:

    g'(y) = \frac{1}{f'(x)} where f'(x) = \frac {1}{\sqrt{x^3+1}}

    however, this expression gives me g'(y) and g''(y) (not x). What can I do to find the g(x)'s ?
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  2. #2
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    Quote Originally Posted by jenkki View Post
    Given that g is the inverse function of f, where:

    f(x) = \int^x_0 \frac {1}{\sqrt{t^3+1}} dt

    for f: [0,\infty) \rightarrow \mathbb{R}


    I have to show that:

    g''(x) = \frac{3}{2} (g(x))^2


    At first I thought that all I had to do was to use the relation:

    g'(y) = \frac{1}{f'(x)} where f'(x) = \frac {1}{\sqrt{x^3+1}}

    however, this expression gives me g'(y) and g''(y) (not x). What can I do to find the g(x)'s ?
    So far, you have found that g'(y) = \sqrt{x^3+1}. Now you need to differentiate again to find g''(y). But remember that this means the second derivative of g with respect to y. So you need to differentiate \sqrt{x^3+1} with respect to y. Use the chain rule for that: differentiate it with respect to x and then multiply the result by \frac{dx}{dy}. You should find that g''(y) = \frac{3x^2}{2\sqrt{x^3+1}}\,\frac{dx}{dy} = \frac32x^2 (because \frac{dx}{dy}  = 1\big/\frac{dy}{dx}). But x = g(y) (because g is the inverse of f), so you get g''(y) = \frac32\bigl(g(y)\bigr)^2. All that remains is to trade the variable y for an x.
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