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# hw4 - Granillo Yvette Homework 4 Due 3:00 am Inst Edward...

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Granillo, Yvette – Homework 4 – Due: Sep 22 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 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) [ ( - 4 , 3) [ (3 , 10) 2. ( - 10 , - 4) [ ( - 4 , 10) 3. ( - 10 , 3) [ (3 , 10) correct 4. ( - 10 , - 4] [ [3 , 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 = - 4 in the case of f and x = - 4 , 3 in the case of g . But the sum of continuous functions is again continuous, so f + g is certainly continuous on ( - 10 , - 4) [ ( - 4 , 3) [ (3 , 10) . The only question is what happens at x 0 = - 4 , 3. 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 = - 4, lim x → - 4 - { f ( x ) + g ( x ) } = - 2 = f ( - 4) + g ( - 4) = lim x → - 4+ { f ( x ) + g ( x ) } , while at x 0 = 3, lim x 3 - { f ( x ) + g ( x ) } = - 7 6 = - 5 = lim x 3+ { f ( x ) + g ( x ) } . Thus, f + g is continuous at x = - 4, but not at x = 3. Consequently, on ( - 10 , 10) the sum f + g is continuous at all x in ( - 10 , 3) [ (3 , 10) . keywords: Stewart5e, 002 (part 1 of 1) 10 points Below is the graph of a function f . 2 4 6 - 2 - 4 - 6 2 4 6 8 - 2 - 4

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Granillo, Yvette – Homework 4 – Due: Sep 22 2005, 3:00 am – Inst: Edward Odell 2 Use the graph to determine all the values of x on ( - 6 , 6) at which f fails to be continuous. 1. x = - 3 2. x = - 3 , 3 correct 3. none of these 4. x = 3 5. no values of x Explanation: Since f ( x ) is defined for all values of x on ( - 6 , 6), the only values of x in ( - 6 , 6) at which the function f is discontinuous are those for which lim x x 0 f ( x ) 6 = f ( x 0 ) or lim x x 0 - f ( x ) 6 = lim x x 0 + f ( x ) . The only possible candidates here are x 0 = - 3 and x 0 = 3. But at x 0 = - 3 f ( - 3) = 9 6 = lim x → - 3 f ( x ) = 6 , while at x 0 = 3 lim x 3 - f ( x ) = 6 6 = lim x 3+ f ( x ) = 4 . Consequently, on ( - 6 , 6) the function f fails to be continuous only at at x = - 3 , 3 . keywords: Stewart5e, 003 (part 1 of 1) 10 points If f and g are continuous functions such that lim x 3 [7 f ( x ) - g ( x )] = 9 , f (3) = 2 , find the value of g (3). 1. g (3) = 14 2. g (3) = 5 correct 3. g (3) = 9 4. g (3) = 2 5. g (3) = 23 Explanation: Since f and g are continuous functions, lim x 3 (7 f ( x ) - g ( x )) = 7 lim x 3 f ( x ) - lim x 3 g ( x ) = 7 f (3) - g (3) = 14 - g (3) . But lim x 3 (7 f ( x ) - g ( x )) = 9 . Consequently, g (3) = 14 - 9 = 5 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Find all values of x at which the function f defined by f ( x ) = x - 6 x 2 - x - 30 is continuous, expressing your answer in in- terval notation. 1. ( -∞ , - 5) ( - 5 , 6) (6 , ) correct 2. ( -∞ , - 5) ( - 5 , ) 3. ( -∞ , - 5) ( - 5 , - 6) ( - 6 , ) 4. ( -∞ , 6) (6 , ) 5. ( -∞ , - 6) ( - 6 , 5) (5 , )
Granillo, Yvette – Homework 4 – Due: Sep 22 2005, 3:00 am – Inst: Edward Odell 3 Explanation: After factorization the denominator be- comes x 2 - x - 30 = ( x - 6)( x + 5) , so f can be rewritten as f ( x ) = x - 6 ( x - 6)( x + 5) = 1 ( x + 5) whenever x 6 = 6. At x = 6 both the numer- ator and denominator will be zero; thus f will not be defined, hence not continuous, at x = 6. Elsewhere f is a ratio of polynomial functions and so will be continuous except at zeros of its denominator. Thus

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