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Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval. y=sin(x³), (1, 2) Let f(x)=sin(x³). Then fis ---Select--- on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select-- Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following. 01<√ [call this value A] < 2 ✓on R. The zeros sin(x) are at x- for n in Z, so we note that 0 <1 << < 2m < 8 < 3m.

Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval. y=sin(x³), (1, 2) Let f(x)=sin(x³). Then fis ---Select--- on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select-- Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following. 01<√ [call this value A] < 2 ✓on R. The zeros sin(x) are at x- for n in Z, so we note that 0 <1 << < 2m < 8 < 3m.

Calculus: Early Transcendentals
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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Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval. y = sin(x³), (1, 2) Let f(x) = sin(x³). Then fis ---Select--- ✓on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select--- ✓on R. The zeros of sin(x) are at x = Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following. 0 1 <i [call this value A] < 2 0 1 <√√/37 [call this value A] <√√/2 0 1 <3 [call this value A] < 2 3 <√√/=/ T 0 1 [call this value A]<√√2 for n in Z, so we note that 0 <1 << < 3/7 < 2π < 8 < 3π.
Transcribed Image Text:Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval. y = sin(x³), (1, 2) Let f(x) = sin(x³). Then fis ---Select--- ✓on the interval [1, 2] since f is the composite of the sine function and the cubing function, both of which are ---Select--- ✓on R. The zeros of sin(x) are at x = Applying the intermediate value theorem and taking the pertinent cube roots of these values in the inequality, we have the following. 0 1 <i [call this value A] < 2 0 1 <√√/37 [call this value A] <√√/2 0 1 <3 [call this value A] < 2 3 <√√/=/ T 0 1 [call this value A]<√√2 for n in Z, so we note that 0 <1 << < 3/7 < 2π < 8 < 3π.
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