Question

How do you find the average rate of change of `g(x)=1/(x-2)` over [0,3]?

Answer 1

I'll use: the average rate of change of function `g` on interval `[a,b]` is `(g(b)-g(a))/(b-a)` and later discuss the mean value of the derivative.

Explanation:

The average rate of change of function `g` on interval `[a,b]` is `(g(b)-g(a))/(b-a)`

But on `[0,3]`, this function has an asymptote. Ignoring this, we get

`(g(3)-g(0))/(3-0) = 1/2`

Interestingly, the rate of change at any point in `[0,3]` is negative.

Here is the graph
graph{1/(x-2) [-0.87, 5.29, -1.211, 1.866]}

It seems odd to me to call this an "average" rate of change.

Indeed if we define the average rate of change to be the mean of the rates of change (the mean value of the derivative), then there is no average rate of change.

`f'(x) = -1/(x-2)^2` is not integrable on `[0,3]` so it has no mean value.