University:
East Georgia State College
Course:
MATH 1540 | Calculus I
Academic year:

2024
Views:

153

Pages:

3
Author:

somospumasl3o

Question 1 In each of the following, list three terms that continue the arithmetic or geometric sequences. Identify the sequences as arithmetic or geometric. Sequences: 2 , 6 , 18 , 54 , 162 2,6,18,54,162 1 , 8 , 15 , 22 , 29 1,8,15,22,29 11 , 15 , 19 , 23 , 27 11,15,19,23,27 Answer: Sequence: 2 , 6 , 18 , 54 , 162 2,6,18,54,162: Ratio of consecutive terms is constant ( 𝑟 = 6 2 = 3 r= 2 6 =3). Sequence type: Geometric, common ratio 𝑟 = 3 r=3. Next three terms: 486 , 1458 , 4374. 486,1458,4374. Sequence: 1 , 8 , 15 , 22 , 29 1,8,15,22,29: Difference of consecutive terms is constant ( 𝑑 = 8 − 1 = 7 d=8−1=7). Sequence type: Arithmetic, common difference 𝑑 = 7 d=7. Next three terms: 36 , 43 , 50. 36,43,50. Sequence: 11 , 15 , 19 , 23 , 27 11,15,19,23,27: Difference of consecutive terms is constant ( 𝑑 = 15 − 11 = 4 d=15−11=4). Sequence type: Arithmetic, common difference 𝑑 = 4 d=4. Next three terms: 31 , 35 , 39. 31,35,39. Question 2 (a) To calculate: The value of 𝑃 ( 2 ) P(2) without a calculator using both forms of the polynomial. The value of 𝑃 ( 2 ) P(2) is 236 236. (b) The least amount of arithmetic operations was performed in (c) using Horner's Method. Answer: (a) Given: A fourth-degree polynomial 𝑃 ( 𝑥 ) = 3 𝑥 4 + 5 𝑥 3 + 4 𝑥 2 + 3 𝑥 + 1 P(x)=3x 4 +5x 3 +4x 2 +3x+1. Rewritten using Horner's Method: 𝑃 ( 𝑥 ) = ( ( ( 3 𝑥 + 5 ) 𝑥 + 4 ) 𝑥 + 3 ) 𝑥 + 1. P(x)=(((3x+5)x+4)x+3)x+1. Calculating 𝑃 ( 2 ) P(2): 𝑃 ( 2 ) = ( ( ( 3 ( 2 ) + 5 ) 2 + 4 ) 2 + 3 ) 2 + 1. P(2)=(((3(2)+5)2+4)2+3)2+1. = ( ( ( 6 + 5 ) 2 + 4 ) 2 + 3 ) 2 + 1. =(((6+5)2+4)2+3)2+1. = ( ( 11 ⋅ 2 + 4 ) ⋅ 2 + 3 ) ⋅ 2 + 1. =((11⋅2+4)⋅2+3)⋅2+1. = ( 26 ⋅ 2 + 3 ) ⋅ 2 + 1. =(26⋅2+3)⋅2+1. = ( 55 ⋅ 2 ) + 1 = 110 + 1 = 236. =(55⋅2)+1=110+1=236. (b) Using Horner's Method minimizes arithmetic operations. Question 3 (a) To calculate: How to rewrite any polynomial without powers of 𝑥 x. (b) To calculate: The relation between the new form and synthetic division. Answer: (a) Rewrite 𝑃 ( 𝑥 ) P(x): 𝑃 ( 𝑥 ) = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛 − 1 𝑥 𝑛 − 1 + … + 𝑎 0 P(x)=a n x n +a n−1 x n−1 +…+a 0 Can be rewritten as: 𝑃 ( 𝑥 ) = ( ( ( 𝑎 𝑛 𝑥 + 𝑎 𝑛 − 1 ) 𝑥 + 𝑎 𝑛 − 2 ) 𝑥 + … + 𝑎 1 ) 𝑥 + 𝑎 0 . P(x)=(((a n x+a n−1 )x+a n−2 )x+…+a 1 )x+a 0 . (b) Synthetic division uses the same arithmetic operations (addition/multiplication) as the rewritten form, allowing efficient evaluation. Question 4 To find: The real zeros of the polynomial function 𝑓 ( 𝑥 ) = 𝑥 4 − 4 𝑥 3 − 4 𝑥 2 + 16 𝑥 f(x)=x 4 −4x 3 −4x 2 +16x. Answer: Rewritten: 𝑓 ( 𝑥 ) = 𝑥 ( 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16 ) . f(x)=x(x 3 −4x 2 −4x+16). Factoring: 𝑓 ( 𝑥 ) = 𝑥 ( 𝑥 + 2 ) 2 ( 𝑥 − 4 ) . f(x)=x(x+2) 2 (x−4). Real zeros: 𝑥 = − 2 , 0 , 2 , 4. x=−2,0,2,4. Question 5 To calculate: Polynomial equation with real coefficients that has roots 1 , 2 − 3 𝑖 1,2−3i. Answer: Roots: 1 , 2 − 3 𝑖 , 2 + 3 𝑖 1,2−3i,2+3i. Polynomial: ( 𝑥 − 1 ) [ ( 𝑥 − ( 2 − 3 𝑖 ) ) ( 𝑥 − ( 2 + 3 𝑖 ) ) ] = 0. (x−1)[(x−(2−3i))(x−(2+3i))]=0. Simplify: ( 𝑥 − 1 ) [ ( 𝑥 − 2 ) 2 + 9 ] = 0. (x−1)[(x−2) 2 +9]=0. Expand: ( 𝑥 − 1 ) ( 𝑥 2 − 4 𝑥 + 13 ) = 0. (x−1)(x 2 −4x+13)=0. Final polynomial: 𝑥 3 − 5 𝑥 2 + 12 𝑥 − 13 = 0. x 3 −5x 2 +12x−13=0.