When ave. rate of change = instantaneous rate of change

[DespicableMe]

Homework Statement

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.

Homework Equations

IROC = [ f(x+h) - f(x) ] /h
AROC = f(x₂) - f(x₁) / x₂ - x₁

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.

 

[Mark44]

Homework Statement

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.

Homework Equations

IROC = [ f(x+h) - f(x) ] /h
AROC = f(x₂) - f(x₁) / x₂ - x₁

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.

You might not have seen the Mean Value Theorem yet, but it guarantees that if f is continuous on some interval [a, b], and differentiable on (a, b), then there exists a number c such that f'(c) = (f(b) - f(a))/(b - a).

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.