2. a. f(x) = (x+2)/(x +4). Use the Mean Value Theorem on the interval [0, 4] to find all values of x in the interval (0, 4) such that the value of the first derivative of f at those points is (f(b) – f(a))/(b – a). b. For the interval [-5, 0], (b) – f(a))/(b – a)=f(0) – f(-5))/(0 – (-5)) = (½ – 3)/(0 + 5) = -2 but, since the first derivative of f(x) is always positive, nowhere on that interval does the first derivative of f(x) = -½. Thus the Mean Value Theorem does not hold for f(x) on this interval. Why?

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...
Question
2. a. f(x) = (x+2)/(x +4). Use the Mean Value Theorem on the interval [0, 4] to find all
values of x in the interval (0, 4) such that the value of the first derivative of f at those
points is (f(b) – f(a))/(b – a).
b. For the interval [-5, 0],
fb) - fa))(b — а)-GG0) — f-5)(0 — (-5))
= (½ – 3)/(0 + 5)
= -½
but, since the first derivative of f(x) is always positive, nowhere on that interval does
the first derivative of f(x) = -½. Thus the Mean Value Theorem does not hold for f(x)
on this interval. Why?
Transcribed Image Text:2. a. f(x) = (x+2)/(x +4). Use the Mean Value Theorem on the interval [0, 4] to find all values of x in the interval (0, 4) such that the value of the first derivative of f at those points is (f(b) – f(a))/(b – a). b. For the interval [-5, 0], fb) - fa))(b — а)-GG0) — f-5)(0 — (-5)) = (½ – 3)/(0 + 5) = -½ but, since the first derivative of f(x) is always positive, nowhere on that interval does the first derivative of f(x) = -½. Thus the Mean Value Theorem does not hold for f(x) on this interval. Why?
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