Please help me with these questions. I am having trouble understanding what to do.Thank you Consider fla$(2)(x³+2x)such that f(x)|< C|x2| whenever x > k.2x+1Arrange the steps in the correct order to show that f(x) is Q(x²) for all x>1.x²+2x*³+2xIt follows that for all x > 1, ≤²²x². Since <0 for x > 1,2x+12x+1we may apply the absolute value without changing the left side: 2x+1ResetTherefore, if x > 1,x³+2x2x+13/2< x².If x> 1, then ½ x² + 1> x² + x² = 3x².Since a fraction gets bigger when the denominator gets bigger,2x+1*³+2x x³+2x2.>=Since a fraction gets smaller when the denominator gets bigger,*³+2x *+2x=+1.2x+12xIt follows that for all x > 1, 12+2x². SirSincex+2x2x+1> 0 for x 1,we may apply the absolute value without changing the left side: x².2x+1Therefore, if x > 1, +*+2*> x².2x+1If x>1, then ½ x²+1 < x²+x² = x². For each of these functions, find the least integer n such that f(x) is O(x).If f(x) = 2x² + x³logx, then the least integer n is

(23+2x) Consider fa 2x+1 such that f(a)|C|r2| whenever a > k. Arrange the steps in the correct order to show that f(x) is Ox²) for all x>1. Reset It follows that for all x > 1, x².s Since x+2x <0 for x > 1, 2x+1 +2x 2x+1 we may apply the absolute value without changing the left side: 2x+1 x+2x 2x+1 Therefore, if x > 1, x². Since a fraction gets bigger when the denominator gets bigger, +2x 2x+1 +2x 2x x²+1.1 Since a fraction gets smaller when the denominator gets bigger, x²+2x x+2x 1 = x²+1. 2x+1 2x It follows that for all x > 1, x². Since +2x 2x+1 >0 for x> 1, 2x+1 we may apply the absolute value without changing the left side: 2x+1 Therefore, if x > 1,>x². 2x+1 If x>1. then x²+1<²x²+x²= x²

(23+2x) Consider fa 2x+1 such that f(a)|C|r2| whenever a > k. Arrange the steps in the correct order to show that f(x) is Ox²) for all x>1. Reset It follows that for all x > 1, x².s Since x+2x <0 for x > 1, 2x+1 +2x 2x+1 we may apply the absolute value without changing the left side: 2x+1 x+2x 2x+1 Therefore, if x > 1, x². Since a fraction gets bigger when the denominator gets bigger, +2x 2x+1 +2x 2x x²+1.1 Since a fraction gets smaller when the denominator gets bigger, x²+2x x+2x 1 = x²+1. 2x+1 2x It follows that for all x > 1, x². Since +2x 2x+1 >0 for x> 1, 2x+1 we may apply the absolute value without changing the left side: 2x+1 Therefore, if x > 1,>x². 2x+1 If x>1. then x²+1<²x²+x²= x²

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.3: Zeros Of Polynomials

Problem 32E

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