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Math Help - Inverses and Second Derivatives

  1. #1
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    Inverses and Second Derivatives

    Problem:
    If f is a one-to-one, twice differentiable function with inverse function g, show that:

    g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )

    and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
    ___________________________

    My work so far:
    For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
    Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
    f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
    (f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3

    I'm not sure where to go from here. What can I do about the second derivatives? Thank you!
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  2. #2
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    Quote Originally Posted by uberbandgeek6 View Post
    Problem:
    If f is a one-to-one, twice differentiable function with inverse function g, show that:

    g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )

    and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
    ___________________________

    My work so far:
    For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
    Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
    f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
    (f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3

    I'm not sure where to go from here. What can I do about the second derivatives? Thank you!

    Part I

    Remember the identity that for inverses, we have

    f[g(x)]=x.

    Taking the derivative, we get

    f'[g(x)]g'(x)=1.

    Divide by f'g to get:

    g'(x)=(f'[g(x)])^{-1}.

    Take the derivative again, yielding

    g''(x)=(-1)(f'[g(x)])^{-2}f''[g(x)]g'(x).

    Remember that g'(x)=(f'[g(x)])^{-1} and substitute:

    g''(x)=(-1)(f'[g(x)])^{-2}f''[g(x)](f'[g(x)])^{-1}.

    And then simplify to yield

    g''(x)=-\frac{f''[g(x)]}{(f'[g(x)])^{3}}.

    Part II

    Now, when we say that f is increasing and concave upward, we are essentially saying that

    f'(t),f''(t)>0 for all t\in D. Since g(x)\in D, then

    f'[g(x)],f''[g(x)]>0.

    We can use that information with our previous equations to infer that g''(x)<0.
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