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Calculus I group study Mod 3

# Calculus I group study Mod 3 - Inflections points are(π 2...

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Calculus I Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0- 618-60624-6). Calculus SG3 Module 4 4A 4.1 #30 y = 3 - | t – 3|, [-1, 5] dy / dt = -1(t-3)(1) / (t-3) by u(u’) / |u| dy / dt = (t - 3) / |3 – t| simplify dy / dt is not applicable at t=3 calculate dy / dt at t = 3 and at the endpoint of the interval [-1, 5] |3 – t| / (t - 3) = |3 – 1| / (1 - 3) = -2 / 2 = -1 t = -1 t = 3, 3 t = 5, 1 Absolute maxima is (3,3) Absolute minima is (-1,-1) 4B 4.4 #26 ƒ(x) = x + 2 cos x [0,2π] First derivative ƒ’(x) = 1 - 2 sin x Second derivative ƒ”(x) = - 2 cos x ƒ”(x) = 0 When cos x = 0 x = π / 2, 3π / 2 0 < x <2π Interval 0 < x <π / 2 π / 2< x <3π / 2 3π / 2 < x <2π Test value x = π / 4 x = 3π / 4 x = 7π / 4 Sign of ƒ”(x) ƒ”(π / 4) = -√2<0 ƒ”(3π / 4) = √2<0 ƒ”(3π / 4) = -√2<0 ƒ(x) is concave downward at (0, π / 2 and (3π / 2 , 2π) concave upward at (π / 2, 3π /2) Concavity changes at x = π / 2 and 3π / 2

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Unformatted text preview: Inflections points are (π / 2, π / 2), (3π / 2, 3π / 2) 4C 4.5 #26 lim [(1 /2)x – 4 / x 2 ] x >∞ lim [(1 /2)x] – lim [4 / x 2 ] x >∞ x >∞ 1 / 2 (-∞) -0 = - ∞ 4D 4.7 #16 ƒ (x) = (x + 1) 2 (x, y) =[x, (x+1) 2 ] Distance ‘d’ between this point and the given (5,3) ƒ(x) x(5, 3) (-1, 0) d = √ (x – 5) 2 + [(x + 1) 2 – 3] 2 d = √ x 4 + 4x 3 + x 2 – 18x + 29 let D = x 4 + 4x 3 + x 2 – 18x + 29 So, dD / dx = 4x 3 + 12x 2 + 2x - 18 dD / dx = (x-1)(4x 2 + 16x + 18) dD / dx = 0 x = 1 or 4x 2 + 16x + 18 = 0 x = 1 or 2 (2x 2 + 8x + 9) = 0 x = 1 or x = -8 ± √(64 – 72) / 4 x = 1 since other values of x are not real When x = 1, ƒ(x) = (x + 1) 2 =4 Therefore the point (x, y) = (1,4) d 2 D / dx 2 = 12x 2 + 24x +2 = + when x = 1 or d = √D is minimum when x = 1 The point (1,4) is closest to the given point (5,3) the closest distance is d = √ (1 – 5) 2 + [(4 - 3) 2 d = √17 ≈ 4.123...
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