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...
Related questions
Question
![Show that cos 3x = 3x has a solution in the interval [0, 1]. Hint: Show that f(x)
= 3x – cos 3x has a zero in [0, 1].
Step 1 of 3
To show that cos 3x = 3x has a solution in the interval [0, 1], we could try to directly solve for some value of x in the interval
[0, 1] that satisfies the equation.
However, this is not easily done. We can make this problem easier by noting that finding an x satisfying cos 3x = 3x is
equivalent to finding an x that satisfies 3x – cos 3x = 0.
We can therefore set f(x) = 3x – cos 3x and look for zeroes of f(x) in the interval [0, 1]. To help prove the existence of a zero
of f(x) in [0, 1], we will use the Intermediate Value Theorem.
The Intermediate Value Theorem states that if f(x) is continuous on a closed interval [a, b] and f(a)
# f(b), then
for every value M between f(a) and f(b), there exists at least one value c E (a, b) such that f(c)
f(b) X
M
As the difference of a polynomial and a trigonometric function, both of which are continuous for all real numbers,
f(x)
= 3x – cos 3x |is
is continuous on the interval [0, 1] and therefore we can
can apply the
Intermediate Value Theorem.
Step 2 of 3
To apply the Intermediate Value Theorem to f(x)
= 3x – cos 3x on [0, 1], evaluate f(0) and f(1). (Round your answers to two
decimal places.)
f(0)
f(1) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9bdd44c0-80af-4006-aeb1-064d54c4c394%2F688a428d-dedf-4c98-a329-e8d25698e07f%2Fgqngg48_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Show that cos 3x = 3x has a solution in the interval [0, 1]. Hint: Show that f(x)
= 3x – cos 3x has a zero in [0, 1].
Step 1 of 3
To show that cos 3x = 3x has a solution in the interval [0, 1], we could try to directly solve for some value of x in the interval
[0, 1] that satisfies the equation.
However, this is not easily done. We can make this problem easier by noting that finding an x satisfying cos 3x = 3x is
equivalent to finding an x that satisfies 3x – cos 3x = 0.
We can therefore set f(x) = 3x – cos 3x and look for zeroes of f(x) in the interval [0, 1]. To help prove the existence of a zero
of f(x) in [0, 1], we will use the Intermediate Value Theorem.
The Intermediate Value Theorem states that if f(x) is continuous on a closed interval [a, b] and f(a)
# f(b), then
for every value M between f(a) and f(b), there exists at least one value c E (a, b) such that f(c)
f(b) X
M
As the difference of a polynomial and a trigonometric function, both of which are continuous for all real numbers,
f(x)
= 3x – cos 3x |is
is continuous on the interval [0, 1] and therefore we can
can apply the
Intermediate Value Theorem.
Step 2 of 3
To apply the Intermediate Value Theorem to f(x)
= 3x – cos 3x on [0, 1], evaluate f(0) and f(1). (Round your answers to two
decimal places.)
f(0)
f(1) =
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