Practice Exam # 2 Math 105 Fall 2013

Practice Exam #2 Math 105 Fall 2013 1. (a) lim(x->7)g(x) = g(7) = -3 The first equality is true because g(x) was assumed to be continuous at x=7. The second equality was given. ** (b). g.f(5) = g(f(5)) = g(7) = -3** Each value was given in assumptions. ** (c) No f(7) ≠ 5 because by assumption, f(x) was** ** assumed** ** to NOT be continuous at x=7, so by** ** definition of continuity.** ** 5 = lim(x->7)f(x) ≠ f(7).** ** given** ** [z(a)y = (5/6)x + x^(5/6) + x^(-5/6) + (5x+6)^(1/2) + (5x+6)^(-1/2)]** ** (5/6)x^(-1/6) - (5/6)x^(-11/6) + (5/2)(5x+6)^(-1/2) (5) - (1/2)(5x+6)^(-3/2) (5)** ** x** ** 2** ** (b)** ** [y = ((2√x + x^2) / (x^(3/3) + (2/3)x))^(2/3)]** ** [2 * (1 + 3x^2 - 2√x + x^3) / (2√x)]^(2/3)** ** 3x** ** 2** ** 2** ** 3** ** x** ** 2** ** 3** ** + x** ** 3** 2(c) f(x) = (3/x^2 - 2/x^3)^9 * sqrt(5-x^2) f'(x) = (3/x^2 - 2/x^3)^9 * (1/2 * (5-x^2)^(-1/2)) * (-2x) + sqrt(5-x^2) * 9 * (3/x^2 - 2/x^3)^8 * (-6x^-3 + 6x^-4) ** (d) y = (1/x - 6x^3) / sqrt(7x + x^8)** ** y' = sqrt(7x+x^8) * (-1/x^2 - 18x^2) - (1/x - 6x^3) * (1/2 * (7x+x^8)^(-1/2)) * (7+8x^7)** ** 7x+x^8** 3. y = (6x + sqrt(8+x^2))^(3/2) y(1) = (6+sqrt(9))^(3/2) = 3^3 = 27 ** y' = (3/2) * (6x + sqrt(8+x^2))^(1/2) * [6 + (1/2 * (8+x^2)^(-1/2)) * (2x)]** ** y'(1) = (3/2) * sqrt(6+sqrt(9))^3 * [6 + (1/sqrt(9))]** ** = (3/2) * sqrt(9)^3 * [18/3 + 1/3]** ** = 27 * 19/3 = 57 specific slope** ** 2** ** Point (1, 27) Slope y'(1) = 57/2** ** Equation of Tangent Line** ** y-27 = (57/2) * (x-1)** ** y = (57/2)x - 57/2** ** y = (57/2)x - 3/2** 4. f(x) = (3x-1)/(2-5x) (1) f'(x) = lim(h->0) [f(x+h) - f(x)]/h = lim(h->0) [(3(x+h)-1)/(2-5(x+h)) - (3x-1)/(2-5x)]/h= lim(h->0) [(2-5x)(3x+3h-1) - (3x-1)(2-5x-5h)] / [(2-5x-5h)(2-5x)] * (1/h)= lim(h->0) [6x+6h-2-15x^2-15xh+5x - (6x-15x^2-15xh-2+5x+5h)] / [(2-5x-5h)(2-5x)] * (1/h)= lim(h->0) [6x+6h-2-15x^2-15xh+5x - 6x+15x^2+15xh+2-5x-5h] / [(2-5x-5h)(2-5x)] * (1/h)= lim(h->0) [h] / [(2-5x-5h)(2-5x)] * (1/h) = 1/(2-5x)^2 (2) f'(x) = [(2-5x)(3) - (3x-1)(-5)] / (2-5x)^2 = (6-15x+15x-5) / (2-5x)^2 = 1/(2-5x)^2 Simplify f'(x) = (2-5x)^-2 => f''(x) = -2(2-5x)^-3(-5) = 10/(2-5x)^3 5. f(x) = (5+3x^2)^8 * (7-x^2)^3 f'(x) = (5+3x^2)^8 * 3(7-x^2)^2 (-2x) + (7-x^2)^3 * 8(5+3x^2)^7 (6x) = x(5+3x^2)^7 * (7-x^2)^2 * 6[(-1)(5+3x^2) + 8(7-x^2)] = x(5+3x^2)^7 * (7-x^2)^2 * 6[-5-3x^2 + 56-8x^2] = x(5+3x^2)^7 * (7-x^2)^2 * 6[-11x^2 + 51] => x = 0 or 5+3x^2 = 0 or 7-x^2 = 0 or -11x^2 + 51 = 0 3x^2 = -5 No Solution x = ±√7 x = ±√(51/11) 6. f(x) = 5x / (1+x) f'(x) = [(1+x)(5) - 5x(1)] / (1+x)^2 = 1 / (1+x)^2 f'(0) = 1 / (1+0)^2 = 1 f'(1) = 1 / (1+1)^2 = 1/4 f'(2) = 1 / (1+2)^2 = 1/9 7.(a) x / (y+1) = x^2 - y^2 d/dx [x / (y+1)] = d/dx [x^2 - y^2] (y+1)(1) - x * dy/dx / (y+1)^2 = 2x - 2y * dy/dx Plug in x=1, y=0 1 - dy/dx = 2 - 0 1 - dy/dx = 2 dy/dx = -1 Specific slope at (1,0) Equation of Tangent Line: y - 0 = -1(x - 1) y = -x + 1 7(b) 4(x+y)^2 = x^2 * y^2 d/dx [4(x+y)^2] = d/dx [x^2 * y^2] 8(x+y)[1 + dy/dx] = x^2 * 2y * dy/dx + y^2 * (2x) Plug in (-2, 1) 8(-2+1)[1 + dy/dx] = (4) * 2 * dy/dx - 4 -8[1 + dy/dx] = 8 * dy/dx - 4 -8 - 8 * dy/dx = 8 * dy/dx - 4 -4 = 16 * dy/dx dy/dx = -1/4 Specific slope at (-2, 1) Equation of Tangent Line: y - 1 = -1/4(x - (-2)) y - 1 = -1/4(x + 2) y - 1 = -1/4x - 1/2 y = -1/4x + 1/2