Answer :
To solve these problems, we'll find the average rate of change of the function over the given interval and compare it with the instantaneous rates of change at the endpoints of the interval.
93. Function: [tex]f(t) = 4t + 5[/tex], Interval: [1, 2]
Average Rate of Change:
The average rate of change of a function [tex]f(t)[/tex] over an interval [tex][a, b][/tex] is given by:
[tex]\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}[/tex]
For [tex]f(t) = 4t + 5[/tex], calculate [tex]f(1)[/tex] and [tex]f(2)[/tex]:[tex]f(1) = 4(1) + 5 = 9[/tex]
[tex]f(2) = 4(2) + 5 = 13[/tex]
Calculate the average rate of change:
[tex]\frac{13 - 9}{2 - 1} = \frac{4}{1} = 4[/tex]Instantaneous Rates of Change:
Since [tex]f(t) = 4t + 5[/tex] is a linear function, its derivative [tex]f'(t) = 4[/tex] is constant. Therefore, the instantaneous rate of change at any point, including the endpoints [tex]t = 1[/tex] and [tex]t = 2[/tex], is always 4, which is the same as the average rate of change.
94. Function: [tex]f(t) = t^2 - 7[/tex], Interval: [3, 3.1]
Average Rate of Change:
Similar to above, calculate:
[tex]\text{Average Rate of Change} = \frac{f(3.1) - f(3)}{3.1 - 3}[/tex]
Calculate [tex]f(3)[/tex] and [tex]f(3.1)[/tex]:[tex]f(3) = 3^2 - 7 = 9 - 7 = 2[/tex]
[tex]f(3.1) = (3.1)^2 - 7 = 9.61 - 7 = 2.61[/tex]
Calculate the average rate of change:
[tex]\frac{2.61 - 2}{3.1 - 3} = \frac{0.61}{0.1} = 6.1[/tex]Instantaneous Rates of Change:
To find the instantaneous rates of change at the endpoints, we need the derivative of [tex]f(t)[/tex]:
[tex]f'(t) = 2t[/tex]Calculating for [tex]t=3[/tex] and [tex]t=3.1[/tex]:
At [tex]t=3[/tex]: [tex]f'(3) = 2(3) = 6[/tex]
At [tex]t=3.1[/tex]: [tex]f'(3.1) = 2(3.1) = 6.2[/tex]
Thus, the instantaneous rates of change at the endpoints are slightly different from the average rate of change, being 6 and 6.2 respectively. The average rate of change calculated for this small interval is 6.1, which is in line between these two instantaneous rates.
