Question

How do you find the average rate of change of `h(x)=x^2+3x-1` over [x, x+h]?

Answer 1

For a function `f`, the average rate of change of `f` over interval `[a,b]` is defined to be `(f(b)-f(a))/(b-a)`

Explanation:

This question uses `h` to mean two different things. To try to avoid confusion, let's rename the function `f`.

`f(x) = x^2+3x-1`

`f(x+h) = (x+h)^2+3(x+h) -1`

The average rate of change is

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

`(f(x+h)-f(x))/h = (overbrace([(x+h)^2+3(x+h) -1])^(f(x+h))-overbrace([x^2+3x-1])^(f(x)))/h`

expand insude the brackets

`= ([x^2+2xh+h^2+3x+3h-1]-[x^2+3x-1])/h`

remove the brackets (distribute the `-` sign)

`= (x^2+2xh+h^2+3x+3h-1-x^2-3x+1)/h`

Simplify the numerator

`= (cancel(color(red)(x^2))+2xh+h^2+cancel(color(green)(3x))+3h-cancel(1)-cancel(color(red)(x^2))-cancel(color(green)(3x))+cancel(1))/h`

`= (2xh+h^2+3h)/h`

factor the `h` in the numerator and reduce the fraction

`= (cancel(h)(2x+h+3))/(cancel(h)1)`

Finish writing

`= 2x+h+3`