A function f(x) is (g(x)) if and only if there are positive constants k, C₁, and C₂ such that C₁lg(x) ≤ f(x)| ≤ C₂lg(x) whene x> K. Click and drag the steps in the correct order to show that 3x² + x + 1 is ©(x²). (You must provide an answer before moving to the next part.) 3x² + x + 1 is 0(3x²) since 3x² is O(3x² + x + 1) since Thus, 3x² + x + 1 is (3x²). 3x² 3x² + x + 1 for x > 0. Take k = 1, C, = 1, and C₂ = 2. Then, 1 3x² ≤ 3x² + x + 1 ≤ 2 3x² for all x > 0. Then, 1 3x² ≤ 3x² + x + 1 ≤2 3x² for all x> 1. 3x² + x + 1 ≤ 3x² + 3x²= 2 3x² for all x>1. Take k = 0, C₁ = 1, and C₂ = 0.

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Q: Let f(x) = 3x + 2 and g(x) = 5x – 8. Find all values of x for which f(x) > g(x).

A: f(x)=3x+2  ,  g(x)=5x-8

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A: given: fx=2x-9g(x)=5x+4 we have to find all values of x for which f(x)>g(x)

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