140Bquiz8csols - 4. f ( x ) = cos x (1 + sin x ) 2 . i)...

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1 MATH 140B QUIZ#8 Solutions Due: Mon Oct 22, 2007 1. i) True . Use the quotient rule. ii) True . Critical points at x = ± 1 . iii) False . A local max occurs at the point (1 , 2). Since f ( - 1) = - 2 , f (1) = 2 , f ( - 3) = - 1 . 2 and f (3) = 1 . 2 , it is also a global max. Therefore, C) is the answer. 2. i) False . Applying the product rule and factoring, you should Fnd that this is wrong by a negative sign. ii) False . The are only 3 critical points at x = 0 , 6 , 8 . iii) True . f (0) = f (8) = 0 and f (6) = 20 . 8 . Therefore, the absolute min occurs occurs at both x = 0 and x = 8 . Therefore, B) is the answer. 3. i) True . Using the quotient rule and simplifying. ii) False . Critical points at x = 0 , 3 π/ 2 . Now, f (0) = 0 , f (2 π ) = 0 , f ( π/ 2) = 1 / 2 , and f (3 π/ 2) = - 1 / 4 . Absolute min occurs at x = 3 π/ 2 . iii) False . Absolute max occurs x = π/ 2 . Therefore, A) is the answer.
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Unformatted text preview: 4. f ( x ) = cos x (1 + sin x ) 2 . i) True . Critical point at x = / 2 only. ii) True . f ( / 2) =-1 / 8 so a local max at x = / 2. Since it the unique critical point in the interval, it is also global. iii) True . As x 3 / 2 , the numerator of f ( x ) oes to-1 and the denominator goes to zero. Therefore, f ( x ) goes to- and there is no abs. min. Therefore, E) is the answer. 5. H ( x ) = (40 + 5 x )(20-2 x ) . i) True . H is quadratic. ii) False . H has negative leading coecient. iii) False . H ( x ) =-10 x 2 + 20 x + 800 and so H ( x ) =-20 x + 20 . Therefore x = 1 . Therefore, A) is the answer....
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This note was uploaded on 03/30/2008 for the course MATH 140B taught by Professor Fabbri,marc during the Fall '07 term at Pennsylvania State University, University Park.

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