S are eventually positive we should use the first

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’s are eventually positive, we should use the first derivative test. f ( i ) = i + 3 i 2 + 10 , i≥ 1 * If f ' ( i ) <0, then f ( i ) will be decreasing. * a. Quotient Rule: i ( ¿¿ 2 + 10 ) 2 f ' ( i ) = ( i 2 + 10 ) 1 −( i + 3 ) 2 i ¿ = 2 i i ( ¿¿ 2 + 10 ) 2 i 2 + 10 ( ¿¿ 2 + 6 i ) ¿ ¿ = i ( ¿¿ 2 + 10 ) 2 i 2 6 i + 10 ¿
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Since the denominator is positive because of the squares inside and outside the parentheses, we will need to look at the numerator to see when it is negative. b. Set i 2 6 i + 10 = 0 The polynomial cannot be factored. However, i can still be determined via the quadratic formula. i = −(− 6 ) ± (− 6 ) 2 4 (− 1 )( 10 ) 2 (− 1 ) ,i≈ 7.359 1.359 Since i is a positive integer in the series and i≥ 1 , we only care about i≈ 1.359 . When plugged back into ( 1.359 ) 2 6 ( 1.359 )+ 10 , this will equal zero. So, we will need to pick a value greater than 1.359 to verify that the series will eventually be nonincreasing. Let’s plug 2 into i 2 6 i + 10 to verify this claim. c. Plug a value greater than 1.359 into i 2 6 i + 10 : ( 2 ) 2 6 ( 2 ) + 10 =− 4 12 + 10 = ¿ -6 Thus, f ' ( i ) <0 when i > 1.359 . 3. u n 0 (Hass & Weir, 2008, p. 611) a. lim i→ ∞ i + 3 i 2 + 10 b. L’Hopital’s Rule: lim i→ ∞ 1 2 i 1 2 lim i→∞ 1 i = 1 2 0 = 0 C. Determine the convergence or divergence of the given infinite series according to the results of the alternating series test. The series converges by the Alternating Series Test.
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D. Hass, J., & Weir, M. D. (2008). Thomas calculus: Early transcendentals . Boston: Pearson Addison-Wesley.
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