High School

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------------------------------------------------ In Exercises 93-96, find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval.

93. f(t) = 4t + 5, [1, 2]
94. f(t) = t² - 7, [3, 3.1]

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]


  1. Calculate [tex]f(t)[/tex] at the endpoints:
    [tex]f(1) = 4(1) + 5 = 9[/tex]
    [tex]f(2) = 4(2) + 5 = 13[/tex]


  2. 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]


  1. 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]


  2. 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]


  3. 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]




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.

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