Question

The function below shows the number of car owners f(t), in thousands, in a city in different years t: f(t) = 0.25t^2 - 0.5t + 3.5 The average rate of change of f(t) from t = 2 to t = 6 is ____ thousand owners per year

Answer 1

Hi there!

• f(6) = (0.25 × 6 - 0.5) × 6 + 3.5 = 9.5
• f(2) = (0.25 × 2 - 0.5) × 2 × 3.5 = 3.5

Th' average rate of change (m) is the ratio of th' change in function value w.r.t width of the Interval.

Acc'rding to rhe question :-

m =
`\frac {f(6) - f(2)}{6 - 2}`

m =
`\frac {9.5 - 3.5}{6 - 2}`

m =
`\frac {6}{4}` =
`\frac {3}{2}` = 1.5

Hence,
The required answer is 1.5 thousand owners per year.

~ Hope it helps!

Answer 2

The average rate of change (m) is the ratio of the change in function value to the width of the interval:
m = (f(6) - f(2))/(6 - 2)

To compute this, we need to compute f(6) and f(2).
f(6) = (0.25*6 -0.5)*6 +3.5 = 9.5
f(2) = (0.25*2 - 0.5)*2 +3.5 = 3.5
Then the average rate of change is
m = (9.5 - 3.5)/(6 - 2) = 6/4 = 1.5

The average rate of change is 1.5 thousand owners per year.