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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].

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].

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.3: The Addition And Subtraction Formulas
Problem 80E
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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) =
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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