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Unformatted text preview: Create assignment, 57950, Final 1, Dec 03 at 4:37 pm 1 This printout should have 106 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC4b24s 50:02, calculus3, multiple choice, > 1 min, wordingvariable. 001 If f is differentiable on ( 2 , 2) and f ( x ) ≤ 3 , ( 2 < x < 2) , f ( 1) = 7 , find the smallest value of M so that the in equality f (1) ≤ M holds for all such f . 1. M = 13 correct 2. M = 16 3. M = 15 4. M = 14 5. M = 12 6. no such M exists Explanation: Since f is differentiable on ( 2 , 2), it is continuous on [ 1 , 1] and differentiable on ( 1 , 1). The MVT applies thus to f on the interval [ 1 , 1]: there exists some c, 1 < c < 1 such that f (1) f ( 1) 2 = f ( c ) . Consequently, if f ( x ) ≤ 3 on ( 1 , 1) and f ( 1) = 7, we see that f (1) = 2 f ( c ) + f ( 1) ≤ 13 = M . StewartC5 04 02 17 50:02, calculus3, multiple choice, > 1 min, wordingvariable. 002 How many real roots does the equation x 5 + 3 x + 2 = 0 have? 1. exactly one real root correct 2. exactly two real roots 3. exactly three real roots 4. exactly four real roots 5. no real roots Explanation: Define a function f by f ( x ) = x 5 + 3 x + 2 . Then the roots of the equation x 5 + 3 x + 2 = 0 are the xintercepts of the graph of f . Now f ( x ) > 0 for x very large, while f ( x ) < 0 for x very large. Thus the graph of f must cross the xaxis at least once. Suppose the graph crosses the x at values x = a and x = b with a < b , i.e. , f ( a ) = f ( b ) = 0 , ( a < b ) . On the other hand, f is a polynomial function, it is continuous and differentiable for all x . Hence Rolle’s Theorem applies, so there exists some c , a < c < b , at which f ( c ) = 0. But f ( x ) = 5 x 4 + 3 > for all x which isn’t consistent with f ( c ) = 0 for some c . Consequently, the graph of f can’t have xintercepts at both x = a and x = b , so the equation has exactly one root . Create assignment, 57950, Final 1, Dec 03 at 4:37 pm 2 Conceivably, the equation could have a re peated root at x = d , say. But then f ( x ) = ( x d ) 2 g ( x ) for some polynomial g in which case f ( d ) = ‡ 2( x d ) g ( x ) + ( x d ) 2 g ( x ) ·fl fl fl x = d = 0 . This again is inconsistent with f ( x ) > 0 for all x , so the equation cannot have repeated roots either. CalC4c05b 50:03, calculus3, multiple choice, < 1 min, wordingvariable. 003 When Sue uses first and second derivatives to analyze a particular continuous function y = f ( x ) she obtains the chart y y y 00 x < 3 + x = 3 4 3 < x < x = 0 1 1 < x < 2 + x = 2 1 DNE x > 2 + + Which of the following can she conclude from her chart?...
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This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas.
 Fall '09
 RAdin

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