Question

Question 5

5 pts
Let f(x)
40e^.06t. What is the average rate of change of f over the
interval [0,3]? Give your answer as a decimal with one place of
accuracy to the right of the decimal point.

Answer 1

`R= 2.6`

Explanation:

Given

`f(x) = 40e^{.06t`

`[a,b] = [0,3]`

Required

Determine the average rate of change (R)

This is calculated as follows:

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

Where

`a = 0` and
`b = 3`

Substitute 0 for a and 3 for b

`R = (f(3) - f(0))/(3 - 0)`

`R = (f(3) - f(0))/(3)`

`R = (1)/(3)[f(3) - f(0)]`

Calculate f(3)

`f(x) = 40e^{.06t`

`f(3) = 40e^(.06*3)`

`f(3) = 40e^(0.18)`

`f(3) = 40 * 1.197`

`f(3) = 47.88`

Calculate f(0)

`f(x) = 40e^{.06t`

`f(0) = 40e^(.06*0)`

`f(0) = 40e^(0)`

`f(0) = 40*1`

`f(0) = 40`

So, the expression:

`R = (1)/(3)[f(3) - f(0)]`

`R= (1)/(3)[47.88 - 40]`

`R= (1)/(3)[7.88]`

`R= 2.62666666667`

`R= 2.6`

Hence, the average rate of change is approximately 2.6