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Show that tan(x) > x for 0 < x < T/2. [Hint: Show that f(x) = tan(x) – x is increasing on (0, A/2).] f(x) = tan(x) – x → f'(x) = > 0 for 0 < x < 을 since > 1 for 0 < x < 곡, Sof is -Select--- (0, 2). Thus, f(x) [? f(0) = 0 for 0 < x < tan(x) – x ? v0 tan(x) > x for on 0 < x < -

Show that tan(x) > x for 0 < x < T/2. [Hint: Show that f(x) = tan(x) – x is increasing on (0, A/2).] f(x) = tan(x) – x → f'(x) = > 0 for 0 < x < 을 since > 1 for 0 < x < 곡, Sof is -Select--- (0, 2). Thus, f(x) [? f(0) = 0 for 0 < x < tan(x) – x ? v0 tan(x) > x for on 0 < x < -

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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**Title: Understanding the Inequality \( \tan(x) > x \) for \( 0 < x < \pi/2 \)** **Objective:** To show that \( \tan(x) > x \) for the interval \( 0 < x < \pi/2 \). **Hint:** Demonstrate that the function \( f(x) = \tan(x) - x \) is increasing on the interval \( (0, \pi/2) \). **Explanation:** 1. **Function Definition:** Let \( f(x) = \tan(x) - x \). 2. **Derivative Calculation:** Compute the derivative, \( f'(x) = \sec^2(x) - 1 \). 3. **Analyzing \( f'(x) \):** - Since \( \sec^2(x) = 1 + \tan^2(x) \), we have \( \sec^2(x) > 1 \) for \( 0 < x < \pi/2 \). - Therefore, \( f'(x) = \sec^2(x) - 1 > 0 \) for \( 0 < x < \pi/2 \). 4. **Conclusion for Monotonicity:** - As \( f'(x) > 0 \), the function \( f(x) = \tan(x) - x \) is increasing on the interval \( (0, \pi/2) \). 5. **Evaluation at a Specific Point:** - Evaluate \( f(x) \) and \( f(0) = 0 \) for \( 0 < x < \pi/2 \), indicating \( f(x) > f(0) \). 6. **Final Conclusion:** - Thus, \( \tan(x) - x > 0 \) implies \( \tan(x) > x \) for \( 0 < x < \pi/2 \). This explanation provides a step-by-step approach to proving the inequality \( \tan(x) > x \) within the specified interval by demonstrating the increasing nature of the associated function \( f(x) = \tan(x) - x \).
Transcribed Image Text:**Title: Understanding the Inequality \( \tan(x) > x \) for \( 0 < x < \pi/2 \)** **Objective:** To show that \( \tan(x) > x \) for the interval \( 0 < x < \pi/2 \). **Hint:** Demonstrate that the function \( f(x) = \tan(x) - x \) is increasing on the interval \( (0, \pi/2) \). **Explanation:** 1. **Function Definition:** Let \( f(x) = \tan(x) - x \). 2. **Derivative Calculation:** Compute the derivative, \( f'(x) = \sec^2(x) - 1 \). 3. **Analyzing \( f'(x) \):** - Since \( \sec^2(x) = 1 + \tan^2(x) \), we have \( \sec^2(x) > 1 \) for \( 0 < x < \pi/2 \). - Therefore, \( f'(x) = \sec^2(x) - 1 > 0 \) for \( 0 < x < \pi/2 \). 4. **Conclusion for Monotonicity:** - As \( f'(x) > 0 \), the function \( f(x) = \tan(x) - x \) is increasing on the interval \( (0, \pi/2) \). 5. **Evaluation at a Specific Point:** - Evaluate \( f(x) \) and \( f(0) = 0 \) for \( 0 < x < \pi/2 \), indicating \( f(x) > f(0) \). 6. **Final Conclusion:** - Thus, \( \tan(x) - x > 0 \) implies \( \tan(x) > x \) for \( 0 < x < \pi/2 \). This explanation provides a step-by-step approach to proving the inequality \( \tan(x) > x \) within the specified interval by demonstrating the increasing nature of the associated function \( f(x) = \tan(x) - x \).
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