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Consider the function x² + 4x + 17. Click and drag statements to prove that x³ is not Q(x² + 4x + 17). Then, x³ 1(x² + 4x + 17) when x = 4. Let x² + 4x +17 ≤ 3x³ for all x> 17 so, x² + 4x + 17 is O(x³), with witnesses C = 3, k = 17. It follows that, x³> x² when x> 4. For x ≤ 3C, clearly impossible for the constant C to satisfy for all large x. It follows that, x < 3C. Therefore, x³ is not O(x² + 4x + 17). However, if x³ were O(x² + 4x + 17), then x³ k.

Consider the function x² + 4x + 17. Click and drag statements to prove that x³ is not Q(x² + 4x + 17). Then, x³ 1(x² + 4x + 17) when x = 4. Let x² + 4x +17 ≤ 3x³ for all x> 17 so, x² + 4x + 17 is O(x³), with witnesses C = 3, k = 17. It follows that, x³> x² when x> 4. For x ≤ 3C, clearly impossible for the constant C to satisfy for all large x. It follows that, x < 3C. Therefore, x³ is not O(x² + 4x + 17). However, if x³ were O(x² + 4x + 17), then x³ k.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.1: Polynomial Functions Of Degree Greater Than
Problem 49E
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Consider the function x² + 4x + 17. Click and drag statements to prove that x³ is not 0(x² + 4x + 17). Reset Then, x³ ≤ C(x² + 4x + 17) ≤ 3Cx² for all sufficiently large x. Consider the constant C= 1 and k = 3, for which x³ > 1(x² + 4x + 17) when x = 4. Let x² + 4x +17 ≤ 3x³ for all x> 17 so, x² + 4x + 17 is O(x³), with witnesses C = 3, k = 17. It follows that, x³>x² when x > 4. For x ≤ 3C, clearly impossible for the constant C to satisfy for all large x. It follows that, x<3C. Therefore, x³ is not O(x² + 4x + 17). However, if x³ were O(x² + 4x + 17), then x³ ≤ C(x² + 4x + 17) whenever x>k.
Transcribed Image Text:Consider the function x² + 4x + 17. Click and drag statements to prove that x³ is not 0(x² + 4x + 17). Reset Then, x³ ≤ C(x² + 4x + 17) ≤ 3Cx² for all sufficiently large x. Consider the constant C= 1 and k = 3, for which x³ > 1(x² + 4x + 17) when x = 4. Let x² + 4x +17 ≤ 3x³ for all x> 17 so, x² + 4x + 17 is O(x³), with witnesses C = 3, k = 17. It follows that, x³>x² when x > 4. For x ≤ 3C, clearly impossible for the constant C to satisfy for all large x. It follows that, x<3C. Therefore, x³ is not O(x² + 4x + 17). However, if x³ were O(x² + 4x + 17), then x³ ≤ C(x² + 4x + 17) whenever x>k.
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