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HW5-solutions - wei(jw35975 HW5 milburn(54685 This...

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wei (jw35975) – HW5 – milburn – (54685) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine lim x 1 b 1 x 2 x 1 x 1 B . 1. limit = 1 correct 2. limit = 1 2 3. limit does not exist 4. limit = 1 3 5. limit = 1 6. limit = 1 3 7. limit = 1 2 Explanation: AFter simplifcation we see that 1 x 2 x 1 x 1 = 1 x x ( x 1) = 1 x For all x n = 1. Thus limit = lim x 1 1 x = 1 . 002 (part 1 of 3) 10.0 points Let F be the Function defned by F ( x ) = x 2 64 | x 8 | . (i) Determine lim x 8 + F ( x ) . 1. limit = 16 2. limit does not exist 3. limit = 8 4. limit = 8 5. limit = 16 correct Explanation: AFter Factorization, x 2 64 | x 8 | = ( x + 8)( x 8) | x 8 | . But, For x > 8, | x 8 | = x 8 . Thus F ( x ) = x + 8 , x > 8 , in which case, by properties oF limits, the right hand limit lim x 8 + F ( x ) = 16 . 003 (part 2 of 3) 10.0 points (ii) Determine lim x 8 - F ( x ) . 1. limit does not exist 2. limit = 16 correct 3. limit = 8 4. limit = 16 5. limit = 8 Explanation: AFter Factorization, x 2 64 | x 8 | = ( x + 8)( x 8) | x 8 | . But, For x < 8, | x 8 | = ( x 8) .
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wei (jw35975) – HW5 – milburn – (54685) 2 Thus F ( x ) = ( x + 8) , x < 8 , in which case, by properties of limits, the left hand limit lim x 8 - F ( x ) = 16 . 004 (part 3 of 3) 10.0 points (iii) Use your results for parts (i) and (ii) to determine lim x 8 F ( x ) . 1. limit = 16 2. limit = 8 3. limit = 8 4. limit = 16 5. limit does not exist correct Explanation: By parts (i) and (ii), lim x 8 + F ( x ) n = lim x 8 - F ( x ) . Consequently, the two-sided limit does not exist . 005 10.0 points Determine lim x 1 x 1 x + 3 2 . 1. limit = 2 2. limit = 4 correct 3. limit = 1 4 4. limit doesn’t exist 5. limit = 1 2 Explanation: After rationalizing the denominator we see that 1 x + 3 2 = x + 3 + 2 ( x + 3) 4 = x + 3 + 2 x 1 . Thus x 1 x + 3 2 = x + 3 + 2 for all x n = 1. Consequently, limit = lim x 1 ( x + 3 + 2) = 4 . 006 10.0 points Determine if lim h 0 f (1 + h ) f (1) h exists when f ( x ) = x 2 + 5 x , and if it does, Fnd its value. 1. limit = 7 correct 2. limit = 11 3. limit does not exist 4. limit = 8 5. limit = 10 6. limit = 9 Explanation: Since f (1 + h ) f (1) = (6 + 7 h + h 2 ) 6 , we see that f (1 + h ) f (1) h = h 2 + 7 h h = h + 7 .
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wei (jw35975) – HW5 – milburn – (54685) 3 On the other hand, lim h 0 ( h + 7) = 7 , by Properties of Limits. Consequently, the given limit exists, and limit = 7 . keywords: limit, deFnition of derivative 007 10.0 points ±unctions f and g are deFned on ( 10 , 10) by their respective graphs in 2 4 6 8 2 4 6 8 4 8 4 8 f g ±ind all values of x where the product, fg , of f and g is continuous, expressing your answer in interval notation. 1. ( 10 , 2) u ( 2 , 10) 2. ( 10 , 10) correct 3. ( 10 , 2) u ( 2 , 4) u (4 , 10) 4. ( 10 , 2] u [4 , 10) 5. ( 10 , 4) u (4 , 10) Explanation: Since f and g are piecewise linear, they are continuous individually on ( 10 , 10) except at their ‘jumps’; i.e. , at x = 4 in the case of f and x = 4 , 2 in the case of g . But the product of continuous functions is again continuous, so is certainly continuous on ( 10 , 2) u ( 2 , 4) u (4 , 10) .
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HW5-solutions - wei(jw35975 HW5 milburn(54685 This...

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