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exam 2 - Version 033 Exam 2 Radin(56520 This print-out...

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Version 033 – Exam 2 – Radin – (56520) 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 if lim x → ∞ x 3 4 x 5 x 3 + 4 x 2 + 5 exists, and if it does, find its value. 1. limit does not exist 2. limit = 0 3. limit = 1 5 4. limit = 2 5 5. limit = 3 5 6. limit = 1 5 correct Explanation: After division by x 3 in both the numerator and denominator, we see that x 3 4 x 5 x 3 4 x 2 + 5 = 1 4 x 2 5 + 4 x + 5 x 3 . On the other hand, lim x → ∞ parenleftBig 1 4 x 2 parenrightBig = 1 , while lim x → ∞ parenleftBig 5 + 4 x + 5 x 3 parenrightBig = 5 . Consequently, lim x → ∞ x 3 4 x 5 x 3 + 4 x 2 + 5 exists and limit = 1 5 . 002 10.0 points Find the rational function f so the figure below is the graph of y = f ( x ) on [ 8 , 8]. -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 2 4 6 2 4 6 2 4 6 2 4 6 1. f ( x ) = x 2 + 3 x + 8 x 2 4 2. f ( x ) = x 2 3 x + 8 x 2 4 3. f ( x ) = x 2 3 x + 8 4 x 2 4. f ( x ) = x 2 + 3 x 8 x 2 4 correct 5. f ( x ) = x 2 + 3 x 8 x 2 6. f ( x ) = x 2 + 3 x 8 x 2 + 4 Explanation: The graph has vertical asymptotes at x = ± 2, so a term x 2 4 must appear in the denominator of f ( x ); in particular, therefore, f ( x ) negationslash = x 2 + 3 x 8 x 2 + 4 and f ( x ) negationslash = x 2 + 3 x 8 x 2 ,
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Version 033 – Exam 2 – Radin – (56520) 2 eliminating two of the possible answers. The graph also has a horizontal asymptote y = 1, so the ratio of coefficients of degree 2 terms in the numerator and denominator of f ( x ) must be 1. Thus f ( x ) negationslash = x 2 3 x + 8 4 x 2 , eliminating a third answer. Hence one of f ( x ) = x 2 + 3 x 8 x 2 4 or f ( x ) = x 2 3 x + 8 x 2 4 or f ( x ) = x 2 + 3 x + 8 x 2 4 must hold. But the y -intercept of the graph is positive, while the y -intercepts of these possi- ble f are f (0) = 2 , or 2 and only one has the property f (0) = 2. Con- sequently, f ( x ) = x 2 + 3 x 8 x 2 4 . 003 10.0 points Determine which of the following graphs could be that of f ( x ) = 3 + 2 x 2 1 + x 2 . 1. 2. 3. 4. correct Explanation: The line y = 2 is a horizontal asymptote because lim x → ±∞ 3 + 2 x 2 1 + x 2 = 2 . On the other hand, f ( x ) = 2 x (1 + x 2 ) 2 . Thus the critical point of f occurs at x = 0. In addition, f is increasing on ( −∞ , 0] and decreasing on [0 , ); in particular, therefore, f has a local maximum at x = 0. The only graph with these properties is
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Version 033 – Exam 2 – Radin – (56520) 3 004 10.0 points If f is the function defined on [ 4 , 4] by f ( x ) = x + | x | − 4 , which of the following properties does f have? A. Absolute minimum at x = 0 . B. Differentiable at x = 0 . 1. B only 2. both of them 3. neither of them 4. A only correct Explanation: Since | x | = braceleftbigg x , x 0 , x , x < 0 , we see that f ( x ) = x + | x |− 4 = braceleftbigg 2 x 4 , x 0 , 4 , x < 0 . Thus on [ 4 , 4] the graph of f is 2 4 2 4 2 4 2 4 Consequently: A. True: by inspection.
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