Find the average rate of change of the function over the given interval.

Question

f(t) = 4t² − 5,      [3, 3.1]

Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval.

left endpoint:

right endpoint:

Answer 1

The average rate of change is just the slope of the line connecting the endpoints. To find slope, we use (y₂ - y₁)/(x₂ - x₁) or, in this case [f(t₂) - f(t₁)]/(t₂ - t₁) = [(4(3.1)² - 5) - (4(3)² - 5)]/(3.1 - 3) = (33.44 - 31)/0.1 = 2.44/.1 = 24.4

The instantaneous rate of change is found by taking the derivative and evaluating it at each endpoint. It's probable that you have only discussed using the limit definition of the derivative so I'll use that here:

Instantaneous rate of change =

for f(t) = 4t² - 5:

f(x + h) = 4(x + h)² - 5 = 4x² + 8xh + 4h² - 5

f(x + h) - f(x) = 4x² + 8xh + 4h² - 5 - (4x² - 5) = 8xh + 4h² = h(8x + 4h)

and [f(x + h) - f(x)]/h = h(8x + 4h)/h = 8x + 4h

and finally, the limit as h approaches zero of 8x + 4h = 8x

For t = 3 (left endpoint), we plug in 3 to get 8(3) = 24

For t = 3.1 (right endpoint) we plug in 3.1 to get 8(3.1) = 24.8