Question

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

Answer 1

`-1/((x-2)(x-2+h))`

Explanation:

The average rate of change is the slope of the secant between:

`(x, g(x))` and `(x+h, g(x+h))`

The slope `m` of a line between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula:

`m = (y_2-y_1)/(x_2-x_1)`

So the slope of the secant is:

`(g(x+h)-g(x))/((x+h)-x) = (g(x+h)-g(x))/h`

So given:

`g(x) = 1/(x-2)`

its average rate of change over `[x. x+h]` is:

`(1/(color(blue)((x+h))-2)-1/(color(blue)(x)-2))/h = ((color(red)(cancel(color(black)(x)))-color(brown)(cancel(color(black)(2))))-(color(red)(cancel(color(black)(x)))+h-color(brown)(cancel(color(black)(2)))))/(h(x-2)(x+h-2))`

`color(white)((1/(((x+h))-2)-1/((x)-2))/h) = (-color(red)(cancel(color(black)(h))))/(color(red)(cancel(color(black)(h)))(x-2)(x+h-2))`

`color(white)((1/(((x+h))-2)-1/((x)-2))/h) = -1/((x-2)(x-2+h))`

In particular, we find:

`lim_(h->0) (g(x+h)-g(x))/h = -1/(x-2)^2`

This is the slope of `g(x)` at `x`