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Click and drag the steps in the correct order to show that 3x² + x + 1 is e(x²). You must provide an answer before moving to the next part.) 3x² + x + 1 is 0(3x²) since 3x² is 0(3x² + x + 1) since Thus, 3x² + x + 1 is 0(3x²). 3x²3x² + x + 1 for x > 0.

Click and drag the steps in the correct order to show that 3x² + x + 1 is e(x²). You must provide an answer before moving to the next part.) 3x² + x + 1 is 0(3x²) since 3x² is 0(3x² + x + 1) since Thus, 3x² + x + 1 is 0(3x²). 3x²3x² + x + 1 for x > 0.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Ch 03 Sec 2 Ex 34 (a) - Proof with Big O** Instruction: Click and drag the steps in the correct order to show that \(3x^2 + x + 1\) is \(\Theta(x^2)\). *(You must provide an answer before moving to the next part.)* **Left Column:** 1. \(3x^2 + x + 1\) is \(O(3x^2)\) since 2. \(3x^2\) is \(O(3x^2 + x + 1)\) since **Right Column:** - Thus, \(3x^2 + x + 1\) is \(\Theta(3x^2)\). - \(3x^2 \leq 3x^2 + x + 1\) for \(x > 0\). - Take \(k = 1\), \(C_1 = 1\), and \(C_2 = 2\). - Take \(k = 0\), \(C_1 = 1\), and \(C_2 = 0\). - Then, \(1 \cdot 3x^2 \leq 3x^2 + x + 1 \leq 2 \cdot 3x^2\) for all \(x > 0\). - \(3x^2 + x + 1 \leq 3x^2 + 3x^2 = 2 \cdot 3x^2\) for all \(x > 1\). - Then, \(1 \cdot 3x^2 \leq 3x^2 + x + 1 \leq 2 \cdot 3x^2\) for all \(x > 1\).
Transcribed Image Text:**Ch 03 Sec 2 Ex 34 (a) - Proof with Big O** Instruction: Click and drag the steps in the correct order to show that \(3x^2 + x + 1\) is \(\Theta(x^2)\). *(You must provide an answer before moving to the next part.)* **Left Column:** 1. \(3x^2 + x + 1\) is \(O(3x^2)\) since 2. \(3x^2\) is \(O(3x^2 + x + 1)\) since **Right Column:** - Thus, \(3x^2 + x + 1\) is \(\Theta(3x^2)\). - \(3x^2 \leq 3x^2 + x + 1\) for \(x > 0\). - Take \(k = 1\), \(C_1 = 1\), and \(C_2 = 2\). - Take \(k = 0\), \(C_1 = 1\), and \(C_2 = 0\). - Then, \(1 \cdot 3x^2 \leq 3x^2 + x + 1 \leq 2 \cdot 3x^2\) for all \(x > 0\). - \(3x^2 + x + 1 \leq 3x^2 + 3x^2 = 2 \cdot 3x^2\) for all \(x > 1\). - Then, \(1 \cdot 3x^2 \leq 3x^2 + x + 1 \leq 2 \cdot 3x^2\) for all \(x > 1\).
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