Question

Find the average rate of change of f(x)=9x^2-2 on the interval of [1,t]

Answer 1

`Rate = 9(t+1)`

Explanation:

Given

`f(x) = 9x^2 - 2`

`Interval = [1,t]`

Required

Determine the average rate of change

This is calculated using:

`Rate = (f(b) - f(a))/(b - a)`

Where

`[a,b] = [1,t]`

So, we have:

`Rate = (f(t) - f(1))/(t - 1)`

Solve for f(t) and f(1)

`f(t) = 9t^2 - 2`

`f(1) = 9*1^2 - 2`

`f(1) = 9 - 2`

`f(1) = 7`

So, we have:

`Rate = (9t^2 - 2 - 7)/(t - 1)`

`Rate = (9t^2 - 9)/(t - 1)`

Factorize:

`Rate = (9(t^2 - 1))/(t - 1)`

Apply difference of two squares

`Rate = (9(t- 1)(t+1))/(t - 1)`

`Rate = 9(t+1)`