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Unformatted text preview: silva (jrs4378) Hw 04 gualdani (56455) 1 This printout should have 21 questions. Multiplechoice 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 ) , 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 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 . . silva (jrs4378) Hw 04 gualdani (56455) 3 Consequently, this function is the only one that is not continuous at x = 3....
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This note was uploaded on 08/29/2010 for the course M 46455 taught by Professor Gualdani during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Gualdani

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