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5 Pages

Hprobs5_sol

Course: MATH 101, Fall 2009
School: East Los Angeles College
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Word Count: 540

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Homework MATH101 Solution Sheet 5 1. (a) Can rewrite the equation as: (x 1)2 y 2 + 2 =1 22 1 which is an ellipse centred at (x, y, ) = (1, 0). Note that at x = 0, 4y 2 = 3 so y = 3 ; at x = 1, y = 1 ; at y = 0, x 1 = 2, so 4 x = 1 or 3 . Use implicit dierentiation to nd derivative: 2(x 1) + 8yy = 0 y = Sketch: 1x 2(x 1) = =0 8y 4y at x = 1 . (b) Can rewrite equation (b) as: x2 (y 1)2 =1 22 32 which is...

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Homework MATH101 Solution Sheet 5 1. (a) Can rewrite the equation as: (x 1)2 y 2 + 2 =1 22 1 which is an ellipse centred at (x, y, ) = (1, 0). Note that at x = 0, 4y 2 = 3 so y = 3 ; at x = 1, y = 1 ; at y = 0, x 1 = 2, so 4 x = 1 or 3 . Use implicit dierentiation to nd derivative: 2(x 1) + 8yy = 0 y = Sketch: 1x 2(x 1) = =0 8y 4y at x = 1 . (b) Can rewrite equation (b) as: x2 (y 1)2 =1 22 32 which is a hyperbola centred at (x, y, ) = (0, 1). Note that at x = 0, there is no real solution for y so the curve does not meet the y-axis; at y = 1, x = 2; at y = 0, 9x2 = 40 so x = 2 10/3; As x , x y1 , 3 2 so there are asymptotes at y =1 3x . 2 Use implicit dierentiation to nd derivative: 18x 8(y 1)y = 0 y = 9x =0 4(y 1) at x = 0 . which is not on the curve. So there are no points on the curve where the slope is zero. Sketch: 2. Find local extrema using f and f : f (x) = 8x3 12x2 + 18x 5 f (x) = 24x2 24x + 18 = 6(4x2 + 4x 3) = 6(2x 1)(2x + 3) f (x) = 48x 24 = 24(2x + 1) . 1 So f = 0 when x = 3 or x = 2 . 2 1 At x = 3 , f = +48 (min.); at x = 2 , f = 48 (max). 2 1 Thus ( 3 , 32) is a local minimum and ( 2 , 0) is a local maximum, 2 1 Point of inection when f = 0 i.e. at ( 2 , 16) . Note that when x = 2, y = 25 and y = 49 when x = 3 so there is a zero between x = 3 and x = 2 (actually at x = 5 as can easily be checked: 2 1 Since x = 2 is a (double) zero of f (x), we know (2x 1)2 = 4x2 4x + 1 is a factor which we can divide out: f (x) = (4x2 4x + 1)(2x 5) , 5 so the third zero is indeed at x = 2 . Hence sketch: 3. Let f (x) = C x2 3 = Ax + B + 2x 4 2x 4 (a) Multiply by across the common denominator: x2 3 = 2Ax2 4Ax + 2Bx 4B + C so: 1 2 x : 1 = 2A A= 2 x1 : 0 = 4A + 2B 2B = 4A = 2 0 x : 3 = 4B + C C = 3 + 4B = 1 . From this we nd f (x) = 1 + (b) f = 1 2 =0 2 (2x 4)2 when (2x 4)2 = 4 , x 1 + . 2 2x 4 i.e. when x = 1 or x = 3. Now f = 8 (2x 4)3 which is 1 at x = 1 and 1 at x = 3. So we conclude (1, 1) is a local maximum and (3, 3) is a local minimum. (c) Note that when x = 0, y = 3 and when y = 0, x2 = 3 so x = 3. 4 There is a vertical asymptote at x = 2 . Since y x + 1 as x , there is an asymptote at 2 y= Hence sketch: x + 1. 2 4. Let f (x) = B C 2x2 + 5x 10 =A+ + . (x 1)(x + 2) x1 x+2 (a) As x , 2 = A + 0 + 0, so A = 2. Multiply across by common denominator: 2x2 + 5x 10 = A(x + 2)(x 1) + B(x + 2) + C(x 1) x=1: 3 = 0 + 3B x = 2 : 12 = 0 + 0 3C . From this we nd B = 1 and C = 4, so f (x) = 2 (b) 1 4 =0 2 (x 1) (x + 2)2 (x + 2)2 = 4(x 1)2 so x + 2 = 2(x 1) f = i.e. when x = 4 or x = 0. f = 2 8 + 3 ...

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