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Derivative Interceptions

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
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Associated Topics:
High School Calculus

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