Rates of Change for a Rational Function

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Zib

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Hello,
Could you help me find out at what point and at what interval the instantaneous rate of change would be equal to the average rate of change for the function f(x)=(x-2)/(x-5) ?

I know that to find the instantaneous rate of change, I can use a centered interval difference quotient; (f(a+h)-f(a))/h.
For the average rate of change, the formula is just Delta Y/Delta X.

I really don't have any idea how to begin, but I know that the point can't be x=5, because that's a vertical asymptote for the graph, and that the function also has a horizontal asymptote at y=1.

Please, how do I go about this?
 
Does the instant have to be the center of the interval?

centered interval difference quotient; (f(a+h)-f(a))/h.
Click to expand...

By the way the CENTERED interval difference quotient is ...
(f(a+h)-f(a-h))/(2h)
And, of course the "instantaneous" value is what you get as h gets close to zero.
 
You have posted in the Advanced Algebra section.
Are you in fact in a Calculus class?

(f(a+h)-f(a-h))/(2h) = ((a+h-2)/(a+h-5)-(a-h-2)/(a-h-5))/(2h)
f'(a) = -3/(a-5)2

You could try to express h(a), but I am not sure that is what your problem is requiring. Also, I expect that it may no be true for any point (unless h=0).
By the way, it is always true for f(x) =x2


Perhaps one example is sufficient, Like at a=0.

Then...
((h-2)/(h-5)-(-h-2)/(-h-5))/(2h) = -3/25
Which is only true as h goes to zero

 
Zib said:
at what point and at what interval the instantaneous rate of change would be equal to the average rate of change for the function f(x)=(x-2)/(x-5)

to find the instantaneous rate of change, I can use a centered interval difference quotient; (f(a+h)-f(a))/h.

For the average rate of change, the formula is just Delta y/Delta X.

how do I go about this?
Click to expand...

Hello Zib:

I'm not sure what your class is doing, but there are easier ways to determine instantaneous rate of change than using difference quotients. (As Bob pointed out, what you posted is not a centered difference quotient. The centered difference quotient is also known as a symmetric difference quotient; you may google it.)

Also, without having worked this exercise myself, it seems as though there are many possible intervals that one could first choose to determine the average change, followed by finding a value x where the instantaneous rate equals the just-found average change.

For example, you could choose the interval [6,10], calculate the average change over that interval, and then use the derivative of function f to find where x equals your result. Or, you could begin with [5,9], and do the same thing. Is this how you understand the exericse?

Is your class studying derivatives (i.e., instantaneous rate-of-change functions)? The difference-quotient approach to determining derivatives involves a limit process. As I mentioned, there are easier ways.

What class are you in? What is your class doing?
 
Thank you for your help. Although it was assigned as homework, my teacher said that there was insufficient information to answer the question, so I don't need to be able to solve it.
 
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