The value of t on the closed interval for which the derivative is equals to the average rate of change is given as follows:
t = 1.75.
How to calculate the value of t?
The function in this problem is given as follows:
v(t) = 4t² - 6t + 2.
The average rate of change over an interval [a,b] is given by the change in the output divided by the change in the input, hence:
r = [f(b) - f(a)]/(b - a).
The interval is [0,3], hence the bounds are given as follows:
a = 0, b = 3.
The numeric value at these bounds is of:
- f(3) = 4(3)² - 6(2) + 2 = 26.
Hence the average rate of change is of:
(26 - 2)/(3 - 0) = 8.
The derivative is given as follows:
v'(t) = 8t - 6.
The equality is then calculated as follows:
v'(t) = 8.
8t - 6 = 8.
8t = 14.
t = 14/8
t = 7/4.
t = 1.75.
More can be learned about the average rate of change of a function at https://brainly.com/question/11627203
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