University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

39

Pages:

3
Author:

Ruby Bates

Question 1 For which positive integers 𝑘 k is the following series convergent? ∑ 𝑛 = 1 ∞ 𝑛 ! ( 𝑘 𝑛 ) 𝑛 n=1 ∑ ∞ (k n ) n n! Answer 1: Use the Ratio Test to determine convergence. 𝑎 𝑛 = 𝑛 ! ( 𝑘 𝑛 ) 𝑛 , 𝑎 𝑛 + 1 = ( 𝑛 + 1 ) ! ( 𝑘 𝑛 + 1 ) 𝑛 + 1 a n = (k n ) n n! ,a n+1 = (k n+1 ) n+1 (n+1)! 𝑎 𝑛 + 1 𝑎 𝑛 = ( 𝑛 + 1 ) ! ( 𝑘 𝑛 + 1 ) 𝑛 + 1 ⋅ ( 𝑘 𝑛 ) 𝑛 𝑛 ! a n a n+1 = (k n+1 ) n+1 (n+1)! ⋅ n! (k n ) n Simplify: 𝑎 𝑛 + 1 𝑎 𝑛 = ( 𝑛 + 1 ) 𝑘 𝑛 + 1 ⋅ ( 𝑘 𝑛 ) 𝑛 ( 𝑘 𝑛 + 1 ) 𝑛 + 1 = ( 𝑛 + 1 ) 𝑘 𝑛 + 1 ⋅ 1 𝑘 𝑛 . a n a n+1 = k n+1 (n+1) ⋅ (k n+1 ) n+1 (k n ) n = k n+1 (n+1) ⋅ k n 1 . For 𝑘 > 2 k>2, the denominator grows faster than the numerator, so the series converges. Answer 2: For 𝑘 ≥ 2 k≥2, use the Comparison Test with a similar convergent series. Hence, the series converges for 𝑘 ≥ 2 k≥2. Question 2 Expand ( 𝑎 − 𝑏 ) 3 (a−b) 3 . Answer: Using the binomial expansion formula: ( 𝑎 − 𝑏 ) 3 = 𝑎 3 − 3 𝑎 2 𝑏 + 3 𝑎 𝑏 2 − 𝑏 3 . (a−b) 3 =a 3 −3a 2 b+3ab 2 −b 3 . Question 3 Multiply ( 𝑥 + ℎ ) 3 (x+h) 3 . Answer 1: Use the binomial expansion formula: ( 𝑥 + ℎ ) 3 = 𝑥 3 + 3 𝑥 2 ℎ + 3 𝑥 ℎ 2 + ℎ 3 . (x+h) 3 =x 3 +3x 2 h+3xh 2 +h 3 . Answer 2: Alternatively, expand step-by-step: ( 𝑥 + ℎ ) 3 = ( 𝑥 + ℎ ) ( 𝑥 + ℎ ) ( 𝑥 + ℎ ) . (x+h) 3 =(x+h)(x+h)(x+h). Expand two terms: ( 𝑥 + ℎ ) ( 𝑥 + ℎ ) = 𝑥 2 + 2 𝑥 ℎ + ℎ 2 . (x+h)(x+h)=x 2 +2xh+h 2 . Then: ( 𝑥 2 + 2 𝑥 ℎ + ℎ 2 ) ( 𝑥 + ℎ ) = 𝑥 3 + 3 𝑥 2 ℎ + 3 𝑥 ℎ 2 + ℎ 3 . (x 2 +2xh+h 2 )(x+h)=x 3 +3x 2 h+3xh 2 +h 3 . Question 4 Show that the curve 𝑦 = 2 𝑒 𝑥 + 3 𝑥 + 5 𝑥 2 y=2e x +3x+5x 2 has no tangent line with slope 2. Answer 1: Find the derivative of 𝑦 y: 𝑦 ′ = 𝑑 𝑑 𝑥 ( 2 𝑒 𝑥 + 3 𝑥 + 5 𝑥 2 ) = 2 𝑒 𝑥 + 3 + 15 𝑥 2 . y ′ = dx d (2e x +3x+5x 2 )=2e x +3+15x 2 . The slope of the tangent line is 𝑦 ′ y ′ . Set 𝑦 ′ = 2 y ′ =2: 2 𝑒 𝑥 + 3 + 15 𝑥 2 = 2. 2e x +3+15x 2 =2. Simplify: 2 𝑒 𝑥 + 15 𝑥 2 + 1 = 0. 2e x +15x 2 +1=0. Observation: 2 𝑒 𝑥 > 0 2e x >0 and 15 𝑥 2 > 0 15x 2 >0, so the sum 2 𝑒 𝑥 + 15 𝑥 2 + 1 > 0 2e x +15x 2 +1>0. This equation has no real solutions. Conclusion: The curve has no tangent line with slope 2. Question 5 Factor the polynomial 9 𝑎 2 − 24 𝑎 + 16 9a 2 −24a+16. Answer: Recognize the expression as a perfect square trinomial: 9 𝑎 2 − 24 𝑎 + 16 = ( 3 𝑎 − 4 ) 2 . 9a 2 −24a+16=(3a−4) 2 . Verification: ( 3 𝑎 − 4 ) 2 = 9 𝑎 2 − 24 𝑎 + 16. (3a−4) 2 =9a 2 −24a+16. Final Answer: 9 𝑎 2 − 24 𝑎 + 16 = ( 3 𝑎 − 4 ) 2 . 9a 2 −24a+16=(3a−4) 2 .