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hw 4 - silva(jrs4378 Hw 04 gualdani(56455 This print-out...

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silva (jrs4378) – Hw 04 – gualdani – (56455) 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 which of the following could be the graph of f near the origin when f ( x ) = x 2 7 x + 10 2 x , x negationslash = 2 , 2 , x = 2 . 1. 2. 3. correct 4. 5. 6. Explanation: Since x 2 7 x + 10 2 x = ( x 2)( x + 5) 2 x = 5 x , for x negationslash = 2, we see that f is linear on ( −∞ , 2) uniondisplay (2 , ) , while lim x 2 f ( x ) = 3 negationslash = f (2) . Thus the graph of f will be a straight line of slope 1, having a hole at x = 2. This eliminates four of the possible graphs. But the two remaining graphs are the same except that in one f (2) > lim x 2 f ( x ) ,
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silva (jrs4378) – Hw 04 – gualdani – (56455) 2 while in the other f (2) < lim x 2 f ( x ) . Consequently, must be the graph of f near the origin. 002 10.0 points Find all values of x at which the function f defined by f ( x ) = x 5 x 2 3 x + 2 is continuous, expressing your answer in in- terval notation. 1. ( −∞ , 1) ( 1 , 2) (2 , ) 2. ( −∞ , 1) (1 , ) 3. ( −∞ , 2) ( 2 , 1) ( 1 , ) 4. ( −∞ , 1) (1 , 2) (2 , ) correct 5. ( −∞ , 2) (2 , ) Explanation: After factorization the denominator be- comes x 2 3 x + 2 = ( x 1)( x 2) , so f can be written as f ( x ) = x 5 ( x 1)( x 2) . Being a rational function, it will be contin- uous everywhere except at the zeros of the denominator since it will not be defined at such points. Thus f is continuous everywhere except at x = 1 and x = 2. Hence it will continuous on ( −∞ , 1) (1 , 2) (2 , ) . 003 10.0 points Determine which (if any) of the following functions is not continuous at x = 3. 1. f ( x ) = 1 | x 1 | x 3 1 2 x < 3 2. f ( x ) = braceleftbigg | x 3 | x negationslash = 3 0 x = 3 3. f ( x ) = 1 x 1 x 3 1 2 x < 3 4. all continuous at x = 3 5. f ( x ) = braceleftBigg 1 x 3 x negationslash = 3 3 x = 3 correct 6. f ( x ) = braceleftBigg 15 2 x 3 x negationslash = 3 5 x = 3 Explanation: A function f will be continuous at x = 3 when f (3) exists and lim x 3 f ( x ) = f (3) . Now f (3) exists for all the functions defined above; in addition, inspection shows that all these functions have the property lim x 3 f ( x ) = f (3) except for f ( x ) = braceleftBigg 1 x 3 x negationslash = 3 3 x = 3 . .
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silva (jrs4378) – Hw 04 – gualdani – (56455) 3 Consequently, this function is the only one that is not continuous at x = 3. 004 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 of the values of x on ( 7 , 7) at which f is discontinuous. 1. none of the other answers 2. no values of x 3. x = 1 4. x = 1 , 1 correct 5. x = 1 Explanation: Since f ( x ) is defined everywhere on ( 7 , 7), the function f will be discontinuous at a point x 0 in ( 7 , 7) if and only if lim x x 0 f ( x ) negationslash = f ( x 0 ) or if lim x x 0 f ( x ) negationslash = lim x x 0 + f ( x ) .
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