exam2 kns

# exam2 kns - Version 042 – Exam 2 – gualdani –(56455 1...

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Unformatted text preview: Version 042 – Exam 2 – gualdani – (56455) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine dy/dx when 2 cos x sin y = 3 . 1. dy dx = cot x cot y 2. dy dx = cot x tan y 3. dy dx = tan xy 4. dy dx = tan x 5. dy dx = tan x tan y correct Explanation: Differentiating implicitly with respect to x we see that 2 braceleftBig cos x cos y dy dx − sin y sin x bracerightBig = 0 . Thus dy dx cos x cos y = sin x sin y . Consequently, dy dx = sin x sin y cos x cos y = tan x tan y . 002 10.0 points Find the slope of the tangent line to the graph of x 3 − 2 y 3 − xy = 0 at the point P ( − 1 , − 1). 1. slope = 3 2 2. slope = − 4 5 3. slope = − 2 3 4. slope = − 5 4 5. slope = 4 5 correct 6. slope = 5 4 Explanation: Differentiating implicitly with respect to x we see that 3 x 2 − 6 y 2 dy dx − y − x dy dx = 0 . Consequently, dy dx = 3 x 2 − y 6 y 2 + x . Hence at P ( − 1 , − 1) slope = dy dx vextendsingle vextendsingle vextendsingle P = 4 5 . 003 10.0 points Find the differential dy when y = 2 + sin x 1 − sin x . 1. dy = sin x (1 − sin x ) 2 dx 2. dy = 3 cos x (1 − sin x ) 2 dx correct 3. dy = − 2 cos x 1 − sin x dx 4. dy = 2 sin x (1 − sin x ) 2 dx 5. dy = − 3 cos x 1 − sin x dx 6. dy = − 3 cos x (1 − sin x ) 2 dx Version 042 – Exam 2 – gualdani – (56455) 2 Explanation: After differentiation of y = 2 + sin x 1 − sin x using the quotient rule we see that dy = (1 − sin x ) cos x + cos x (2 + sin x ) (1 − sin x ) 2 dx . Consequently, dy = 3 cos x (1 − sin x ) 2 dx . 004 10.0 points If f is the function whose graph is given by 2 4 6 2 4 6 which of the following properties does f have? A. critical point at x = 2 , B. local minimum at x = 4 , C. f ′ ( x ) > 0 on ( − 1 , 2) . 1. B and C only 2. B only 3. A only correct 4. none of them 5. A and C only 6. all of them 7. C only 8. A and B only Explanation: The given graph has a removable disconti- nuity at x = 4 and a critical point at x = 2. On the other hand, recall that f has a local maximum at a point c when f ( x ) ≤ f ( c ) for all x near c . Thus f could have a local max- imum even if the graph of f has a removable discontinuity at c ; similarly, the definition of local minimum allows the graph of f to have a local minimum at a removable disconitu- ity. So it makes sense to ask if f has a local extremum at x = 4. Inspection of the graph now shows of the three properties A. f has , B. f does not have , C. f does not have . 005 10.0 points Find the absolute maximum value of f ( x ) = 4 − 3 cos 2 x on [ − π, π ]. 1. abs maximum value = 7 2. abs maximum value = 0 3. abs maximum value = 4 correct 4. abs maximum value = 2 5. abs maximum value = 1 6. abs maximum value = 8 Explanation: The absolute maximum value of f on [ − π, π ] occurs (a) either at an endpoint x = − π or x = π , (b) or at a critical point of f in (...
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exam2 kns - Version 042 – Exam 2 – gualdani –(56455 1...

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