Question

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

Answer 1

Average rate change over the interval `[0,1/2]` is `2/15`
Instantaneous rate of change at point `x` is `1/(x-3)^2`

Explanation:

`y=f(x)= -1/(x-3)= -(x-3)^-1`

Average rate of change over the interval `[0,1/2]` is

`R_a= (f(b)- f(a))/(b-a) ;a= 0 , b= 1/2`

`f(1/2)= -1/(1/2-3)= 2/5 ; f(0)= -1/(0-3)= 1/3`

`:. R_a= (2/5-1/3)/(1/2-0)=1/15/1/2=2/15`

Average rate change over the interval `[0,1/2]` is `2/15`

Instantaneous rate of change is derivative of `f(x)` at the

point `x :. f^'(x)=-.-(x-3)^-2= 1/(x-3)^2`

Instantaneous rate of change at point `x` is `1/(x-3)^2` [Ans]