3 calculus Qs(2)

35)http://i1099.photobucket.com/albums/g395/jasoncube/385581db.png 37)http://i1099.photobucket.com/albums/g395/jasoncube/13e9ef9e.png 38)a) y = x³/(x² - 3) = x³/[(x - √3)(x + √3)] = x + 3x/(x² - 3)∴ Asymptotes are y = - √3, y = - √3 and y = xb) y = x³/(x² - 3) ⇒ dy/dx = [(x² - 3)(3x²) - x³(2x)]/(x² - 3)²= (x⁴ - 9x²)/(x² - 3)² = x²(x² - 9)/(x² - 3)c) When Y is maximum orminimum, dy/dx = 0dy/dx = 0 ⇒ x²(x² - 9)/(x² - 3) = 0 ⇒ x = - 3 or0 or 3When - ∞< x < - 3, dy/dx > 0. When - 3< x < - √3, dy/dx < 0∴ Maximum point = (- 3, (- 3)³/[(- 3)² - 3])= (- 3, - 9/2)When - √3 < x < -0, dy/dx > 0. When 0 < x < √3, dy/dx > 0∴ Point of inflexion = (0, 0)When √3 < x < 3, dy/dx < 0. When 3 < x < ∞, dy/dx < 0∴ Minimum point = (3, 3³/(3² - 3)) = (- 3, 9/2)d) By themaximum point, point of inflexion, minimum point and asymptotes.(See:www.wolframalpha.com/input/?i=x%C2%B3%2F%28x%C2%B2+-+3%29)e) x³ - kx² + 3k = 0 ⇒ k = x³/(x² - 3)By d), when - 9/2< k < 9/2, (*) have only 1 real root.(It's because if horizontal lines are drawnwithin - 9/2 < y < 9/2,each line with only interesect the curve x³ - x²y + 3y = 0 at 1 point) I HOPE THIS CAN HELP YOU!~PLEASEFEEL FREE AGAIN TO ASK IF YOU STILL HAVE ANYPROBLEM! ^_^

2012-07-18 15:40:21 補充：
Exceeded the restriction on number of words~

2012-07-18 17:12:11 補充：
The area will be maximum if x = 40/(π + 20).
But there's the restriction 11/(π - 2) ≤ x ≤ 14/(π - 2),
so the nearer to 40/(π + 20) the x is, the nearer to the maximum the area will be.
Within 11/(π - 2) ≤ x ≤ 14/(π - 2), 11/(π - 2) is the nearest to 40/(π + 20),

2012-07-18 17:13:17 補充：
the area is maximum when x = 11/(π - 2).
From c) i), that is H = 3m.