Let ƒ be a function that is differentiable on [a, b] . we defined the average rate of change of ƒ over [a, b] to be ƒ(b) - ƒ(a)/( b - a) and the instantaneous rate of change of ƒ at x to be ƒ′(x). In this chapter we defined the average value of a function. For the new definition of average to be consistent with the old one, we should have ƒ(b) - ƒ(a)/( b - a) = average value of ƒ′ on [a, b]. Is this the case? Give reasons for your answer.
Let ƒ be a function that is differentiable on [a, b] . we defined the average rate of change of ƒ over [a, b] to be ƒ(b) - ƒ(a)/( b - a) and the instantaneous rate of change of ƒ at x to be ƒ′(x). In this chapter we defined the average value of a function. For the new definition of average to be consistent with the old one, we should have ƒ(b) - ƒ(a)/( b - a) = average value of ƒ′ on [a, b]. Is this the case? Give reasons for your answer.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Let ƒ be a function that is differentiable on [a, b] . we defined the average rate of change of ƒ over [a, b] to be
ƒ(b) - ƒ(a)/( b - a) and the instantaneous rate of change of ƒ at x to be ƒ′(x). In this chapter we defined the average value of a function. For the new definition of average to be consistent with the old one, we should have
ƒ(b) - ƒ(a)/( b - a) = average value of ƒ′ on [a, b].
Is this the case? Give reasons for your answer.
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