a) You need to remember the average rate of change formula such that:

`(dy)/(dx) = (f(2pi/3) - f(pi/2))/(2pi/3 - pi/2)`

You need to evaluate `f(2pi/3), ` hence you need to substitute `2pi/3` for theta in `f(theta) = 3cos(theta - pi/6)+ 1` such that:

`f(2pi/3) = 3cos(2pi/3 - pi/6)+ 1 =gt f(2pi/3) = 3cos(3pi/6) + 1` (you need to bring the terms in brackets to a common denominator)

`f(2pi/3) = 3cos(pi/2) + 1 `

You need to substitute 0 for cos `pi/2 ` such that:

`f(2pi/3) =3*0 + 1 =gt f(2pi/3) = 1 `

You need to evaluate `f(pi/32), ` hence you need to substitute `pi/2` for theta in `f(theta) = 3cos(theta - pi/6)+ 1` such that:

`f(pi/2) = 3cos(pi/2 - pi/6)+ 1`

`f(pi/2) = 3cos(2pi/6)+ 1`

`f(pi/2) = 3cos(pi/3)+ 1`

`f(pi/2) = 3(1/2)+ 1 =gt f(pi/2) = 3/2+ 1 = 5/2`

You need to evaluate the average rate of change such that:

`(dy)/(dx) = (1 - 5/2)/(2pi/3 - pi/2)`

`(dy)/(dx) = (-3/2)/(4pi/6 - 3pi/6)`

`(dy)/(dx) = (-3/2)/(pi/6) =gt (dy)/(dx) =-18/(2pi)`

`(dy)/(dx) =-9/pi`

**Hence, evaluating the average rate of change of function between `pi/2` and `2pi/3` yields `(dy)/(dx) =-9/pi` .**

**b) **You need to evaluate the instantaneous rate of change at `theta = 2pi/3` , hence you need to evaluate`f'(2pi/3), ` but first you need to find `f'(theta)` such that:

`f'(theta) = -3sin(theta - pi/6)*(theta - pi/6)'`

`f'(theta) = -3sin(theta - pi/6)`

You need to substitute `2pi/3` for theta in `f'(theta)` such that:

`f'(2pi/3) = -3sin(2pi/3 - pi/6)`

You need to bring the terms inside brackets to a common denominator such that:

`f'(2pi/3) = -3sin(4pi/6 - pi/6)`

`f'(2pi/3) = -3sin(3pi/6) =gt f'(2pi/3) = -3sin(pi/2)`

`f'(2pi/3) = -3 (sin pi/2 = 1)`

**Hence, evaluating the instantaneous rate of change at `theta = 2pi/3` yields `f'(2pi/3) = -3` .**

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.