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Quest HW 3

# Quest HW 3 - roberts(eer474 Quest HW 3 seckin(56425 This...

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roberts (eer474) – Quest HW 3 – seckin – (56425) 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 Functions f and g are defined on ( 10 , 10) by their respective graphs in 2 4 6 8 2 4 6 8 4 8 4 8 f g Find all values of x where the sum, f + g , of f and g is continuous, expressing your answer in interval notation. 1. ( 10 , 4] uniondisplay [2 , 10) 2. ( 10 , 4) uniondisplay ( 4 , 10) correct 3. ( 10 , 4) uniondisplay ( 4 , 2) uniondisplay (2 , 10) 4. ( 10 , 2) uniondisplay (2 , 10) 5. ( 10 , 10) Explanation: Since f and g are piecewise linear, they are continuous individually on ( 10 , 10) except at their ‘jumps’, i.e. , at x = 2 in the case of f and x = 2 , 4 in the case of g . But the sum of continuous functions is again continuous, so f + g is certainly continuous on ( 10 , 4) uniondisplay ( 4 , 2) uniondisplay (2 , 10) . The only question is what happens at x 0 = 2 , 4. To do that we have to check if lim x x 0 { f ( x ) + g ( x ) } = f ( x 0 ) + g ( x 0 ) = lim x x 0 + { f ( x ) + g ( x ) } . Now at x 0 = 2, lim x 2 { f ( x ) + g ( x ) } = 2 = f (2) + g (2) = lim x 2+ { f ( x ) + g ( x ) } , while at x 0 = 4, lim x → − 4 { f ( x ) + g ( x ) } = 4 negationslash = 6 = lim x → − 4+ { f ( x ) + g ( x ) } . Thus, f + g is continuous at x = 2, but not at x = 4. Consequently, on ( 10 , 10) the sum f + g is continuous at all x in ( 10 , 4) uniondisplay ( 4 , 10) . 002 10.0 points If the function f is continuous everywhere and f ( x ) = x 2 16 x + 4 when x negationslash = 4, find the value of f ( 4). 1. f ( 4) = 16 2. f ( 4) = 4 3. f ( 4) = 16 4. f ( 4) = 8 5. f ( 4) = 8 correct 6. f ( 4) = 4

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roberts (eer474) – Quest HW 3 – seckin – (56425) 2 Explanation: Since f is continuous at x = 4, f ( 4) = lim x → − 4 f ( x ) . But, after factorization, x 2 16 x + 4 = ( x 4)( x + 4) x + 4 = x 4 , whenever x negationslash = 4. Thus f ( x ) = x 4 for all x negationslash = 4. Consequently, f ( 4) = lim x → − 4 ( x 4) = 8 . 003 10.0 points Below is the graph of a function f . -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 2 4 6 2 4 6 2 4 2 4 Use this graph to determine all the values of x at which f fails to be continuous on ( 8 , 8). 1. x = 2 , 6 2. x = 6 , 6 3. x = 6 , 2 , 6 correct 4. f is continuous everywhere 5. x = 6 , 2 Explanation: The function f is continuous at a point a in ( 8 , 8) when (i) f ( a ) is defined, (ii) lim x a f ( x ) exists, and (iii) lim x a f ( x ) = f ( a ). We check where one or more of these condi- tions fails. (i) This fails at a = 6. The only other possible candidates are a = 6 and x 0 = 2: (ii) At x 0 = 6 lim x →− 6 f ( x ) = 5 negationslash = lim x →− 6+ f ( x ) = 2 so the limit does not exist; while (iii) at x 0 = 2, f ( 2) = 1 negationslash = lim x → − 2 f ( x ) = 2 , so the limit exists but does not have value f ( a ). Consequently, f fails to be continuous only at x = 6 , 2 , 6 on ( 8 , 8).
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Quest HW 3 - roberts(eer474 Quest HW 3 seckin(56425 This...

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