On which interval does the function f(x) = r+ sin r-1 have a real root? O. Зт 2T O. Зт T, [0.5] None of the listed answers. 21

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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On which interval does the function f(x) = r+ sin r-1 have a real root?
O.
Зт
2T
O.
Зт
T,
[0.5]
None of the listed answers.
21
Transcribed Image Text:On which interval does the function f(x) = r+ sin r-1 have a real root? O. Зт 2T O. Зт T, [0.5] None of the listed answers. 21
Expert Solution
Step 1

We will apply the intermediate value theorem. For a continuous function f(x) if f(a) and f(b) differ in sign then there must be at least one root of f(x) in the interval [a, b]

Step 2

Let's test each option one by one and sequentially:

f(x) = x + sinx - 1

First option: 

f(3π/2) = 3π/2 + sin(3π/2) - 1 = 3π/2 - 1 - 1 > 0

f(2π) = 2π + sin2π - 1 = 2π + 0 - 1 > 0

the function has the same sign at both the boundary points, hence there may not be a solution in this interval.

Second option:

f(π) = π + sinπ - 1 = π - 1 - 1 > 0

f(3π/2) = 3π/2 + sin(3π/2) - 1 = 3π/2 - 1 - 1 > 0

the function has the same sign at both the boundary points, hence there may not be a solution in this interval.

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