HW04-solutions - vasquez(mpv244 HW04 Schultz(56445 This...

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vasquez (mpv244) – HW04 – Schultz – (56445) 1 This print-out should have 23 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 product, fg , of f and g is continuous, expressing your answer in interval notation. 1. ( 10 , 4] uniondisplay [2 , 10) 2. ( 10 , 4) uniondisplay ( 4 , 10) 3. ( 10 , 10) correct 4. ( 10 , 4) uniondisplay ( 4 , 2) uniondisplay (2 , 10) 5. ( 10 , 2) uniondisplay (2 , 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 product of continuous functions is again continuous, so fg 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 ) } = 8 = f (2) g (2) = lim x 2+ { f ( x ) g ( x ) } , while at x 0 = 4, lim x → − 4 { f ( x ) g ( x ) } = 0 = f ( 4) g ( 4) = lim x → − 4+ { f ( x ) g ( x ) } . Thus, fg is continuous at x = 2 and at x = 4. Consequently, the product fg is continuous at all x in ( 10 , 10) . 002 10.0 points If the function f is continuous everywhere and f ( x ) = x 2 25 x + 5 when x negationslash = 5, find the value of f ( 5). 1. f ( 5) = 10 correct 2. f ( 5) = 25 3. f ( 5) = 25 4. f ( 5) = 5 5. f ( 5) = 5 6. f ( 5) = 10 Explanation:
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vasquez (mpv244) – HW04 – Schultz – (56445) 2 Since f is continuous at x = 5, f ( 5) = lim x → − 5 f ( x ) . But, after factorization, x 2 25 x + 5 = ( x 5)( x + 5) x + 5 = x 5 , whenever x negationslash = 5. Thus f ( x ) = x 5 for all x negationslash = 5. Consequently, f ( 5) = lim x → − 5 ( x 5) = 10 . 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 = 4 , 1 , 5 correct 2. f is continuous everywhere 3. x = 4 , 5 4. x = 1 , 5 5. x = 4 , 1 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 = 5. The only other possible candidates are a = 4 and x 0 = 1: (ii) At x 0 = 4 lim x →− 4 f ( x ) = 5 negationslash = lim x →− 4+ f ( x ) = 2 so the limit does not exist; while (iii) at x 0 = 1, f (1) = 1 negationslash = lim x 1 f ( x ) = 2 , so the limit exists but does not have value f ( a ). Consequently, f fails to be continuous only at x = 4 , 1 , 5 on ( 8 , 8). 004 10.0 points Find all values of x at which the function f defined by f ( x ) = x 1 x 2 6 x + 8 is continuous, expressing your answer in in- terval notation.
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