Analysis of Signs of Derivatives of an Inverse FunctionDate: 06/22/2005 at 09:37:35 From: Tommy Subject: The inverse of a concave function I want to be able to reach conclusions about the first and second order derivatives of an inverse function. I have a concave function which I know has f'> 0 and f''< 0. It is a continuous function and can be differentiated as many times as one would like. The function has an inverse, since the Jakobi determinant is not zero. So now I would like to take the inverse of the function f', and be able to conclude what signs the derivaties of this inverse function have. I know that f''< 0 and since f'> 0 and hence cannot take negative values, f'''> 0 has to hold. Now my difficulties arise. Can one conclude anything about the derivatives of the inverse of a function (f') with these characteristics? Date: 06/23/2005 at 17:55:51 From: Doctor Vogler Subject: Re: The inverse of a concave function Hi Tommy, Thanks for writing to Dr. Math. Recall that an inverse function is defined to be a function g such that f(g(x)) = x and g(f(x)) = x for all values x. So we can take the derivative of the first equation and get f'(g(x))g'(x) = 1, which means that g'(x) = 1/f'(g(x)). Since f' is always positive, so too g' is always positive. Then we can take more derivatives of this too: g"(x) = [g'(x)]' = [1/f'(g(x))]' = -[f'(g(x))]'/f'(g(x))^2 = -[f"(g(x))g'(x)]/f'(g(x))^2 = -f"(g(x))/f'(g(x))^3. Now since f' is always positive, and f" is always negative, this means that g" is always positive. And so on. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 06/27/2005 at 06:22:02 From: Tommy Subject: Thank you (The inverse of a concave function) Dear Dr. Vogler, Thank you so much for your reply. It was the definition of an inverse function that had escaped me. Your help is much appreciated! Regards, Tommy |
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