On this question I plugged in f(0) = cos(0) - (0)^3 = 1. Likewise, I did the same with f(1) in my calculator to get -0.4596976941. I tried replicating the process of question 57E here. Do I need to do something different here? Consider the following.cos(x) = x3(a) Prove that the equation has at least one real root.The equation cos(x) = x³ is equivalent to the equation f(x) = cos(x) – x = 0. f(x) is continuous on the interval [0, 1], f(0) =| 1, andf(1) =-.4596976941. Since --Select--- v < 0 < --Select-- vthere is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is aroot of the equation cos(x) = x³, in the interval (0, 1).(b) Use your calculator to find an interval of length 0.01 that contains a root. (Enter your answer using interval notation. Round your answers to two decimal places.)

Answered: Consider the following. cos(x) = x3 (a)…

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 this question I plugged in f(0) = cos(0) - (0)^3 = 1. Likewise, I did the same with f(1) in my calculator to get -0.4596976941. I tried replicating the process of question 57E here. Do I need to do something different here?

Consider the following.
cos(x) = x3
(a) Prove that the equation has at least one real root.
The equation cos(x) = x³ is equivalent to the equation f(x) = cos(x) – x = 0. f(x) is continuous on the interval [0, 1], f(0) =| 1
, and
f(1) =-.4596976941
. Since --Select--- v < 0 < --Select-- v
there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a
root of the equation cos(x) = x³, in the interval (0, 1).
(b) Use your calculator to find an interval of length 0.01 that contains a root. (Enter your answer using interval notation. Round your answers to two decimal places.)

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