buckley (jdb4687) - HW06 - beshaj - (53521) This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. 001 003 2 sin x 1 6x lim x /6 lim x0 27 4 1. limit = 2. limit doesn't exist 3. limit = 25 4 1 2. limit = 2 3 4. limit = 13 2 1 5. limit = 6 2 3 3 4. limit = 2 3 5. limit = 2 004 10.0 points Find the derivative of f when f (x) = 4 tan(x) + 2 cot(x) . 1 6. limit = 3 002 sec2 5x 1 4x2 if the limit exists. 1 1. limit = 3 3. limit = 10.0 points Find the value of 10.0 points Use the definition of the derivative to determine the value of 1 1. f (x) = 4 6 sin2 (x) sin2 (x) cos2 (x) 2. f (x) = 4 6 cos2 (x) sin2 (x) cos2 (x) 3. f (x) = 4 + 6 cos(x) sin(x) cos(x) 4. f (x) = 4 6 cos(x) sin(x) cos(x) 5. f (x) = 4 + 6 cos2 (x) sin2 (x) cos2 (x) 6. f (x) = 4 + 6 sin2 (x) sin2 (x) cos2 (x) 10.0 points Find an equation for the\u0010tangent line to the \u0010 \u0011\u0011 when ,f graph of f at the point P 4 4 f (x) = 4 tan(x) . \u0010 \u0011 1. y = 5x + 11 1 2 \u0010 \u0011 2. y = 3x + 2 1 4 \u0010 \u0011 3. y = 9x + 6 1 4 \u0010 \u0011 4. y = 2x + 14 1 2 \u0010 \u0011 5. y = 8x + 4 1 2 005 Find the value of \u0010 lim sin 4x x0 10.0 points 1 \u0011 2 . sin 3x cos 2x buckley (jdb4687) - HW06 - beshaj - (53521) 2 1. limit = 8 3 5. f (x) = x2 + 2x 3 (1 + x)2 2. limit = 8 5 6. f (x) = 3x (x + 1)3 3. limit does not exist 008 Find f (x) when 4. limit = 2 10.0 points Find the derivative of f when f (x) = 4x cos 2x 3 sin 2x . 1. f (x) = 8x sin 2x 2 cos 2x 2. f (x) = 8x sin 2x 2 cos 2x 3. f (x) = 6 cos 2x + 8x sin 2x 1. f (x) = 5. f (x) = 8x sin 2x 6 cos 2x 10.0 points Find the derivative of f (x) = 1. f (x) = 2. f (x) = (1 + x)2 . 1x x2 2x 3 (1 x)2 3+x (1 x)3 x2 + 2x . p 1 (x + 1) x2 + 2x 2 2(x + 1) 2. f (x) = x2 + 2x x+1 x2 + 2x p 4. f (x) = (x + 1) x2 + 2x 3. f (x) = x+1 2 x2 + 2x p 6. f (x) = 2(x + 1) x2 + 2x 5. f (x) = 4. f (x) = 6 cos 2x 2x sin 2x 007 p f (x) = 4 5. limit = 3 006 10.0 points 009 10.0 points Let f (x) = (3x 1)4 (4x + 4)2 . Find the x-intercept of the tangent line to the graph of f at the point where the graph of f crosses the y-axis. 1. x-intercept = 2. x-intercept = 1 10 1 14 3. x-intercept = 1 5 3. f (x) = x3 (1 + x)3 4. x-intercept = 1 5 4. f (x) = 3 + 2x x2 (1 x)2 5. x-intercept = 1 10 buckley (jdb4687) - HW06 - beshaj - (53521) 010 10.0 points 1. y = Find the value of F (3) when F (x) = f (g(x)) and g(3) = 2, g (3) = 2 , f (3) = 3, f (2) = 6 . 1. F (3) = 8 2. y = 3 y + 3 + 2x x 3 + 2x y x 3. y = y + 3 + 4x x 4. y = y + 3 + 2x x 5. y = y + 3 + 4x x 2. F (3) = 9 6. y = (y + 3 + 4x) 3. F (3) = 10 013 4. F (3) = 11 If y = y(x) is defined implicitly by y 2 + xy + 2 = 0, 5. F (3) = 12 011 10.0 points Find the derivative of H when H(x) = (f (x))2 (g(x))2 and f (x) = 8 g(x) , g (x) = 5 f (x) . 1. H (x) = 13f (x) g(x) 2. H (x) = 26f (x) g(x) 3. H (x) = 13(f (x) + g(x)) 4. H (x) = 6f (x) g(x) 5. H (x) = 3f (x) g(x) 6. H (x) = 3(f (x) + g(x)) 012 10.0 points find the value of dy/dx at the point (3, 1). dy = 3 1. dx (3, 1) dy = 2 2. dx (3, 1) dy 3. = 3 dx (3, 1) dy 4. = 1 dx (3, 1) dy = 2 5. dx (3, 1) dy 6. = 1 dx (3, 1) 014 10.0 points Determine dy/dx when 10.0 points 4 cos x sin y = 7 . Find y when xy + 3x + 2x2 = 1 . 1. dy = cot x tan y dx buckley (jdb4687) - HW06 - beshaj - (53521) dy dx dy 3. dx dy 4. dx dy 5. dx 2. 017 = tan x 4 10.0 points Differentiate the function = tan xy f (x) = cos(ln 4x) . = tan x tan y = cot x cot y 015 10.0 points Find the equation of the tangent line to the graph of 5y 2 xy 4 = 0, at the point P = (1, 1). 1. f (x) = 2. f (x) = 3. f (x) = 5. f (x) = 2. y = x sin(ln 4 x) x 4 sin(ln 4x) x 4. f (x) = 1. 8y = x + 7 4 sin(ln 4x) x 1 cos(ln 4 x) 6. f (x) = sin(ln 4 x) 3. 8y + x = 7 018 10.0 points Find the slope of the line tangent to the graph of ln(xy) + 2x = 0 4. 9y + x = 8 5. 9y = x + 8 016 sin(ln 4 x) x at the point where x = 1. 10.0 points Find the derivative of f when f () = ln (cos 2) . 1. slope = e2 1 2. slope = e2 2 3. slope = 1. f () = 2 tan 2 1 2. f () = sin 2 1 2 e 2 1 4. slope = e2 2 5. slope = e2 3. f () = 2 cot 2 4. f () = 2 cos 2 6. slope = e2 019 10.0 points 5. f () = cot 2 6. f () = 2 tan 2 Determine the value of f (1) when f (x) = 4 ln (2x + 1) . buckley (jdb4687) - HW06 - beshaj - (53521) 1. f (1) = 32 9 2. f (1) = 32 9 3. f (1) = 16 3 4. f (1) = 16 9 5. f (1) = 16 3 6. f (1) = 16 9 020 2! x 5. 3! x 022 10.0 points The following functions are continuous for all x: F1 (x) = cos |(x)| , F2 (x) = sin |(x)| , 2 |x | , x 6= 0 . F3 (x) = x 0, x=0 10.0 points Find the derivative of f (t) = 1. f (t) = 4. 5 ln t . 3 + ln t 3 t(3 + ln t)2 3 ln t 2. f (t) = 3 + ln t Which are differentiable for all x? 1. F1 only 2. F2 only 3. F1 and F2 only 4. all of them 5. F2 and F3 only 3 ln t 3. f (t) = (3 + ln t)2 4. f (t) = ln t 3 + ln t 6. F1 and F3 only 5. f (t) = 1 t(3 + ln t)2 7. F3 only 6. f (t) = 1 t(3 + ln t) 8. none of them 021 10.0 points \u0001 d4 Find 4 x3 ln x . dx 3! 1. 4 ln x x 023 10.0 points Find the derivative of f when n o f (x) = x 7 sin(ln x) + 5 cos(ln x) . 2. 2! x4 1. f (x) = x{2 sin(ln x) + 12 cos(ln x)} 3. 3! x4 2. f (x) = 12 sin(ln x) + 2 cos(ln x) buckley (jdb4687) - HW06 - beshaj - (53521) 3. f (x) = 2 sin(ln x) + 12 cos(ln x) 4. difference = $205.69 4. f (x) = x{12 sin(ln x) + 2 cos(ln x)} 5. difference = $204.69 5. f (x) = 2 sin(ln x) 12 cos(ln x) 024 10.0 points Find the derivative of f (x) = 2x sin 4x + 1 cos 4x . 2 1. f (x) = 8x cos 4x 4 sin 4x 2. f (x) = 8x cos 4x + 4 sin 4x 3. f (x) = 8 cos 4x 4. f (x) = 8x cos 4x 5. f (x) = 8x cos 4x 025 10.0 points Find the value of f (0) when 027 (part 1 of 3) 10.0 points Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant temperature of 190 C. Being a mathematician, she knows that the temperature T of the pie after t minutes of baking will be given by T (t) = 190 Aekt , where A and k are constants. After 17 minutes of baking she notices that the temperature of the pie is 114 C, while after 34 minutes it is 152 C. Determine the constants A and k. 1. T (t) = (190 152 2t/17 ) C 2. T (t) = (190 152 2t/34 ) C f (x) = (1 7x)2 . 3. T (t) = (190 76 4t/17 ) C 026 4. T (t) = (190 152 2t/17 ) C 10.0 points Your aunt who died recently left you $13000. Because of uncertainty in the stock market you decide to invest the money for the next 7 years in Money Market Funds. Both Fund B0 and Fund B1 pay the same annual nominal interest rate of 8%, but Fund B0 compounds monthly, while Fund B1 compounds semi-annually. How much more will your investment be worth if you put the money in Fund B0 rather than Fund B1 ? 5. T (t) = (190 76 2t/34 ) C 028 (part 2 of 3) 10.0 points If the pie is cooked when it reaches a temperature of 171 C, how long does it have to be in the oven? 1. cooking time = 56 minutes 1. difference = $206.69 2. cooking time = 41 minutes 2. difference = $203.69 3. difference = $202.69 6 3. cooking time = 51 minutes buckley (jdb4687) - HW06 - beshaj - (53521) 4. cooking time = 46 minutes 2. Caus 0.125 5. cooking time = 61 minutes 3. Caus 0.113 029 (part 3 of 3) 10.0 points What was the initial temperature of the pie? 1. initial temp = 38 C 2. initial temp = 42 C 3. initial temp = 40 C 4. initial temp = 41 C 7 4. Caus 0.116 5. Caus 0.119 032 10.0 points The amount, $A, in a Wells Fargo savings account satisfies the differential equation dA = 0.08 A. dt Katy deposits $100 in such a savings account. Find out how much money will be in her account after 5 years, leaving the answer in exponential form. 1. A(5) = $100 e40 5. initial temp = 39 C 2. A(5) = $100 e4 030 10.0 points The population of Dime Box grows exponentially. In 1990 the population was 600, and in 2000 it was 1150. Estimate the population in 1997. 1. population in 1997 956 2. population in 1997 986 3. population in 1997 966 4. population in 1997 946 5. population in 1997 976 031 10.0 points A recently discovered radioactive substance Austinium is thought to have fast decay. If scientists find that 48% of the radioactivity remains after 6 years, estimate the value of the decay constant, Caus , for Austinium. 1. Caus 0.122 3. A(5) = $100 e0.4 4. A(5) = $100 e4 5. A(5) = $100 e0.4