Date: 02/10/98 at 00:29:04 From: Eric Arroa Subject: Derivative Interceptions I'm trying to figure out if there is a way to determine the points at which a derivative intersects a function. In the case of a function in the form of x^n, it works out to be f'(n), since nx^(n-1) = x^n when x is n in f'(x). Intuitively, this seems to have a calculus solution (without resorting to mapping between vector maps with multi-index matrices), although I've been a whole hour on it and have found little. I'm not very good at math... Thank you for your help. Eric Arora
Date: 02/10/98 at 11:51:44 From: Doctor Rob Subject: Re: Derivative Interceptions Thanks for writing, Eric. You clearly are better at math than you think, since you say you're not very good at math, yet have invented some of the calculus! The problem of when f'(x) = f(x) is not one commonly seen. Usually the former is a rate of change, and the latter is a value. For example, if x were time and f(x) distance, then f'(x) would be velocity. It would be an unusual problem to ask when the velocity in cm/sec equals the distance in cm. That said, if f(x) is specified in advance, the only way to solve the problem that I know of is to take the derivative f'(x), compute f(x)-f'(x), and find the roots of this new function. In the case of f(x) = x^n, you are interested in the roots of x^n - n*x^(n-1) = x^(n-1)*(x - n) = 0. They are 0 (n-1 times over) and n. By the way, there is a function f(x) such that f(x) = f'(x) for *every* value of x. It is C*e^x, for any constant C. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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