Answer :
To find the average rate of change of a function over a given interval, we use the formula:
[tex]\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}[/tex]
where [tex]f(a)[/tex] and [tex]f(b)[/tex] are the function values at the endpoints [tex]a[/tex] and [tex]b[/tex] of the interval.
Problem 93:
Function: [tex]f(t) = 4t + 5[/tex]
Interval: [tex][1, 2][/tex]
Calculate [tex]f(t)[/tex] at the endpoints:
[tex]f(1) = 4(1) + 5 = 9[/tex]
[tex]f(2) = 4(2) + 5 = 13[/tex]Apply the average rate of change formula:
[tex]\text{Average rate of change} = \frac{13 - 9}{2 - 1} = \frac{4}{1} = 4[/tex]
Since the function [tex]f(t) = 4t + 5[/tex] is linear, the average rate of change over any interval is equal to the instantaneous rate of change (slope), which is 4 at any point.
Problem 94:
Function: [tex]f(t) = t^2 - 7[/tex]
Interval: [tex][3, 3.1][/tex]
Calculate [tex]f(t)[/tex] at the endpoints:
[tex]f(3) = 3^2 - 7 = 9 - 7 = 2[/tex]
[tex]f(3.1) = (3.1)^2 - 7 = 9.61 - 7 = 2.61[/tex]Apply the average rate of change formula:
[tex]\text{Average rate of change} = \frac{2.61 - 2}{3.1 - 3} = \frac{0.61}{0.1} = 6.1[/tex]Find the instantaneous rates of change at the endpoints:
The instantaneous rate of change is given by the derivative [tex]f'(t)[/tex]:
[tex]f'(t) = 2t[/tex]- At [tex]t = 3[/tex]:
[tex]f'(3) = 2 \times 3 = 6[/tex] - At [tex]t = 3.1[/tex]:
[tex]f'(3.1) = 2 \times 3.1 = 6.2[/tex]
- At [tex]t = 3[/tex]:
The average rate of change over the interval [3, 3.1] is 6.1, which is closer to the instantaneous rate of change at 3.1 (6.2) than at 3 (6). This is because the interval is short and the function is a quadratic curve, which slightly increases the rate as [tex]t[/tex] increases.
