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

2024
Views:

402

Pages:

3
Author:

Ruby Bates

Question 1 Linear and quadratic approximation: Linear approximation: 𝑝 1 ( 𝑥 ) = 𝑓 ( 𝑎 ) + 𝑓 ′ ( 𝑎 ) ( 𝑥 − 𝑎 ) p 1 (x)=f(a)+f ′ (a)(x−a) Substitute 𝑓 ( 𝑥 ) = 𝑥 1 3 , 𝑎 = 8 f(x)=x 3 1 ,a=8: 𝑓 ( 8 ) = 8 1 3 = 2 , 𝑓 ′ ( 𝑥 ) = 1 3 𝑥 − 2 3 ⟹ 𝑓 ′ ( 8 ) = 1 3 ⋅ 1 4 = 1 12 f(8)=8 3 1 =2,f ′ (x)= 3 1 x − 3 2 ⟹f ′ (8)= 3 1 ⋅ 4 1 = 12 1 𝑝 1 ( 𝑥 ) = 2 + 1 12 ( 𝑥 − 8 ) p 1 (x)=2+ 12 1 (x−8) Quadratic approximation: 𝑝 2 ( 𝑥 ) = 𝑓 ( 𝑎 ) + 𝑓 ′ ( 𝑎 ) ( 𝑥 − 𝑎 ) + 1 2 𝑓 ′ ′ ( 𝑎 ) ( 𝑥 − 𝑎 ) 2 p 2 (x)=f(a)+f ′ (a)(x−a)+ 2 1 f ′′ (a)(x−a) 2 𝑓 ′ ′ ( 𝑥 ) = − 2 9 𝑥 − 5 3 ⟹ 𝑓 ′ ′ ( 8 ) = − 2 9 ⋅ 1 32 = − 1 144 f ′′ (x)=− 9 2 x − 3 5 ⟹f ′′ (8)=− 9 2 ⋅ 32 1 =− 144 1 𝑝 2 ( 𝑥 ) = 2 + 1 12 ( 𝑥 − 8 ) − 1 288 ( 𝑥 − 8 ) 2 p 2 (x)=2+ 12 1 (x−8)− 288 1 (x−8) 2 Approximation at 𝑥 = 7.5 x=7.5: Linear: 𝑝 1 ( 7.5 ) = 2 + 1 12 ( 7.5 − 8 ) = 2 − 0.5 12 = 1.9583 p 1 (7.5)=2+ 12 1 (7.5−8)=2− 12 0.5 =1.9583 Quadratic: 𝑝 2 ( 7.5 ) = 2 + 1 12 ( 7.5 − 8 ) − 1 288 ( 7.5 − 8 ) 2 = 2 − 0.5 12 − 0.25 288 = 1.95747 p 2 (7.5)=2+ 12 1 (7.5−8)− 288 1 (7.5−8) 2 =2− 12 0.5 − 288 0.25 =1.95747 Question 2 Find the cubic polynomial with roots − 5 , 1 , 𝑖 −5,1,i. Answer If 𝑖 i is a root, then − 𝑖 −i is also a root. Write the factors as: ( 𝑥 + 5 ) , ( 𝑥 − 1 ) , ( 𝑥 2 + 1 ) (x+5),(x−1),(x 2 +1) Multiply: ( 𝑥 + 5 ) ( 𝑥 − 1 ) ( 𝑥 2 + 1 ) = ( 𝑥 2 + 4 𝑥 − 5 ) ( 𝑥 2 + 1 ) (x+5)(x−1)(x 2 +1)=(x 2 +4x−5)(x 2 +1) = 𝑥 4 + 4 𝑥 3 − 5 𝑥 2 + 𝑥 2 + 4 𝑥 − 5 = 𝑥 3 + 3 𝑥 2 − 8 𝑥 + 10 =x 4 +4x 3 −5x 2 +x 2 +4x−5=x 3 +3x 2 −8x+10 Final Answer: 𝑥 3 + 3 𝑥 2 − 8 𝑥 + 10 x 3 +3x 2 −8x+10 Question 3 Factor 2 𝑥 2 + 𝑥 − 1 2x 2 +x−1: Answer 1 Multiply the coefficient of 𝑥 2 x 2 with the constant: 2 ⋅ − 1 = − 2 2⋅−1=−2 Split the middle term: 2 𝑥 2 + 2 𝑥 − 𝑥 − 1 = 2 𝑥 ( 𝑥 + 1 ) − 1 ( 𝑥 + 1 ) = ( 2 𝑥 − 1 ) ( 𝑥 + 1 ) 2x 2 +2x−x−1=2x(x+1)−1(x+1)=(2x−1)(x+1) Question 4 Multiply ( − 2 𝑥 − 3 𝑦 ) ( 3 𝑥 + 2 𝑦 ) (−2x−3y)(3x+2y): Answer Using FOIL: ( − 2 𝑥 − 3 𝑦 ) ( 3 𝑥 + 2 𝑦 ) = − 6 𝑥 2 − 4 𝑥 𝑦 − 9 𝑥 𝑦 − 6 𝑦 2 = − 6 𝑥 2 − 13 𝑥 𝑦 − 6 𝑦 2 (−2x−3y)(3x+2y)=−6x 2 −4xy−9xy−6y 2 =−6x 2 −13xy−6y 2 Question 5 Factor 𝑥 3 + 1 x 3 +1: Answer Using the sum of cubes formula: 𝑎 3 + 𝑏 3 = ( 𝑎 + 𝑏 ) ( 𝑎 2 − 𝑎 𝑏 + 𝑏 2 ) a 3 +b 3 =(a+b)(a 2 −ab+b 2 ) 𝑥 3 + 1 3 = ( 𝑥 + 1 ) ( 𝑥 2 − 𝑥 ⋅ 1 + 1 2 ) = ( 𝑥 + 1 ) ( 𝑥 2 − 𝑥 + 1 ) x 3 +1 3 =(x+1)(x 2 −x⋅1+1 2 )=(x+1)(x 2 −x+1)