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

2024
Views:

118

Pages:

4
Author:

Ruby Bates

Question 1 Determine whether the function 𝑓 ( 𝑥 ) = 𝑥 3 + 2 𝑥 − 4 f(x)=x 3 +2x−4 has a root on [ 1 , 2 ] [1,2] using the intermediate value theorem. If yes, find the root to five decimal places. Answer Step 1: Verify continuity and sign change on the interval [ 1 , 2 ] [1,2]. Plug 𝑥 = 1 x=1: 𝑓 ( 1 ) = 1 3 + 2 ( 1 ) − 4 = − 1 f(1)=1 3 +2(1)−4=−1 Plug 𝑥 = 2 x=2: 𝑓 ( 2 ) = 2 3 + 2 ( 2 ) − 4 = 8 f(2)=2 3 +2(2)−4=8 Since 𝑓 ( 1 ) < 0 f(1)<0 and 𝑓 ( 2 ) > 0 f(2)>0, there is a root in [ 1 , 2 ] [1,2]. Step 2: Find the root using numerical methods: The root is approximately 𝑥 = 1.17951 x=1.17951. Question 2 Write an equation for a rational function with: Vertical asymptotes at 𝑥 = − 2 x=−2 and 𝑥 = − 4 x=−4, 𝑥 x-intercepts at 𝑥 = 1 x=1 and 𝑥 = 3 x=3, Horizontal asymptote at 𝑦 = 8 y=8. Answer Numerator: 𝑓 ( 𝑥 ) = ( 𝑥 − 1 ) ( 𝑥 − 3 ) f(x)=(x−1)(x−3), based on 𝑥 x-intercepts. Denominator: 𝑛 ( 𝑥 ) = ( 𝑥 + 2 ) ( 𝑥 + 4 ) n(x)=(x+2)(x+4), based on vertical asymptotes. Horizontal asymptote 𝑦 = 8 y=8: The leading coefficients must give a ratio of 8. The rational function is: 𝑅 ( 𝑥 ) = 64 ( 𝑥 − 1 ) ( 𝑥 − 3 ) 3 ( 𝑥 + 2 ) ( 𝑥 + 4 ) . R(x)= 3(x+2)(x+4) 64(x−1)(x−3) . Question 3 Find the intervals of increasing and decreasing for 𝑦 = − 2 𝑥 2 + 4 𝑥 + 3 y=−2x 2 +4x+3 using the first derivative. Answer First derivative: 𝑓 ′ ( 𝑥 ) = − 4 𝑥 + 4 f ′ (x)=−4x+4 Solve 𝑓 ′ ( 𝑥 ) = 0 f ′ (x)=0: − 4 𝑥 + 4 = 0 ⟹ 𝑥 = 1 −4x+4=0⟹x=1 Test intervals: For 𝑥 < 1 x<1: Pick 𝑥 = 0 x=0, 𝑓 ′ ( 0 ) = 4 > 0 f ′ (0)=4>0 (increasing). For 𝑥 > 1 x>1: Pick 𝑥 = 2 x=2, 𝑓 ′ ( 2 ) = − 4 < 0 f ′ (2)=−4<0 (decreasing). Result: Increasing on ( − ∞ , 1 ) (−∞,1). Decreasing on ( 1 , ∞ ) (1,∞). Question 4 What determines the end behavior of a polynomial function? Answer The end behavior of a polynomial function is determined by: Degree of the polynomial: Even degree: Both ends go in the same direction. Odd degree: Ends go in opposite directions. Leading coefficient: Positive: Ends rise. Negative: Ends fall. Explanation: For 𝑓 ( 𝑥 ) = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛 − 1 𝑥 𝑛 − 1 + ⋯ + 𝑎 0 f(x)=a n x n +a n−1 x n−1 +⋯+a 0 : The term 𝑎 𝑛 𝑥 𝑛 a n x n dominates for large ∣ 𝑥 ∣ ∣x∣. Thus, the leading coefficient determines the end behavior.