|Jan19-11, 10:32 AM||#1|
When ave. rate of change = instantaneous rate of change
1. The problem statement, all variables and given/known data
Given the function f(x)= (x-2) / (x-5), determine an interval and a point where the ave. R.O.C and the instantaneous R.O.C are equal.
2. Relevant equations
IROC = [ f(x+h) - f(x) ] /h
AROC = f(x2) - f(x1) / x2 - x1
3. The attempt at a solution
I know that in order to satisfy this, the x's must satisfy both equations when using the AROC and IROC formulas.
I'm not sure where to start, though.
I could do guess and check, but I need help for an algebraic method.
|Jan19-11, 12:30 PM||#2|
If you haven't graphed your function, by all means do so. Your function has a vertical asymptote at x = 5, but is continuous everywhere else; i.e., on (-infinity, 5) or (5, infinity). Pick any two numbers that are in the interior of either of these intervals and calculate the average rate of change. Then set f'(x) equal to this number to solve for x.
|Similar Threads for: When ave. rate of change = instantaneous rate of change|
|Instantaneous rate of change (two capacitors)||Introductory Physics Homework||9|
|How to get instantaneous rate of change||Precalculus Mathematics Homework||6|
|instantaneous rate of change of a sphere||Precalculus Mathematics Homework||2|
|instantaneous rate of change||Calculus & Beyond Homework||8|
|Derivative as an Instantaneous rate of change||Calculus & Beyond Homework||8|