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# solution_pd9f - vasquez(mpv244 HW09 Schultz(56445 This...

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vasquez (mpv244) – HW09 – Schultz – (56445) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if the function f ( x ) = x x + 12 satisfies the hypotheses of Rolle’s Theorem on the interval [ 12 , 0], and if it does, find all numbers c satisfying the conclusion of that theorem. 1. c = 8 correct 2. c = 9 3. c = 8 , 9 4. c = 5 5. c = 8 , 8 6. hypotheses not satisfied Explanation: Rolle’s Theorem says that if f is 1. continuous on [ a, b ] , 2. differentiable on ( a, b ) , and 3. f ( a ) = f ( b ) = 0, then there exists at least one c , a < c < b , such that f ( c ) = 0. Now the given function f ( x ) = x x + 12 , is defined for all x ≥ − 12, is continuous on [ 12 , ), and differentiable on ( 12 , ). In addition f ( 12) = f (0) = 0 . In particular, therefore, Rolle’s theorem ap- plies to f on [ 12 , 0]. On the other hand, by the Product and Chain Rules, f ( x ) = x + 12+ x 2 x + 12 = 3 x + 24 2 x + 12 . Thus there exists c, 12 < c < 0, such that f ( c ) = 3 c + 24 2 c + 12 = 0 , in which case c = 8 . 002 10.0 points Let f be a function defined on [0 , 1] such that f (0) = 1 , f (1) = 2 . Consider the following properties that f might have: A. f ( x ) = | 3 x 1 | ; B. f is cont. on [0 , 1] , ; C. f ( x ) = braceleftBigg x 2 + 1 , x negationslash = 1 / 2, 1 , x = 1 / 2 ; which properties ensure that f ( c ) = 1 for some c in (0 , 1)? 1. A and B only 2. B only 3. C only 4. A only 5. none of them correct 6. B and C only 7. all of them 8. A and C only Explanation: The Mean Value Theorem (MVT) says:

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vasquez (mpv244) – HW09 – Schultz – (56445) 2 If f is a function continuous on [ a, b ] and differentiable on ( a, b ) , then f ( c ) = f ( b ) f ( a ) b a for at least one c in ( a, b ) . If one or more of the hypotheses fail, then there need not exist any such c . A. No c : f ( x ) = braceleftBigg 3 , x < 1 / 3, 3 , x > 1 / 3. B. No c : f ( x ) = braceleftBigg 1 , 0 x < 1 / 2 2 x, 1 / 2 x 1. C. No c : ( f ( x ) = 2 x, x negationslash = 1 / 2). 003 10.0 points How many real roots does the equation x 5 + 5 x + 3 = 0 have? 1. exactly four real roots 2. exactly two real roots 3. exactly three real roots 4. exactly one real root correct 5. no real roots Explanation: Define f by f ( x ) = x 5 + 5 x + 3 . Then the roots of the equation x 5 + 5 x + 3 = 0 are the x -intercepts of the graph of f . Now f ( x ) x 5 for | x | large, so f ( x ) → ∞ as x → ∞ , while f ( x ) → −∞ as x → −∞ . Thus the graph of f must cross the x -axis 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 + 5 > 0 for all x which isn’t consistent with f ( c ) = 0 for some c . Consequently, the graph of f can’t have x -intercepts at both x = a and x = b , so the equation has exactly one root .
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solution_pd9f - vasquez(mpv244 HW09 Schultz(56445 This...

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