HW 5 Calc - garcia (jjg2564) Hw 04 gualdani (56410) 1 This...

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Unformatted text preview: garcia (jjg2564) Hw 04 gualdani (56410) 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 , 4 , x = 2 . 1. 2. 3. 4. correct 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 ) , garcia (jjg2564) Hw 04 gualdani (56410) 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 5 x + 6 is continuous, expressing your answer in in- terval notation. 1. ( , 2) (2 , 3) (3 , ) correct 2. ( , 2) ( 2 , 3) (3 , ) 3. ( , 2) (2 , ) 4. ( , 3) (3 , ) 5. ( , 3) ( 3 , 2) ( 2 , ) Explanation: After factorization the denominator be- comes x 2 5 x + 6 = ( x 2)( x 3) , so f can be written as f ( x ) = x 5 ( x 2)( x 3) . 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 = 2 and x = 3. Hence it will continuous on ( , 2) (2 , 3) (3 , ) . 003 10.0 points Determine which (if any) of the following functions is not continuous at x = 5. 1. f ( x ) = 1 | x 3 | x 5 1 2 x < 5 2. all continuous at x = 5 3. f ( x ) = 1 x 3 x 5 1 2 x < 5 4. f ( x ) = braceleftBigg 20 2 x 5 x negationslash = 5 4 x = 5 5. f ( x ) = braceleftbigg | x 5 | x negationslash = 5 x = 5 6. f ( x ) = braceleftBigg 1 x 5 x negationslash = 5 5 x = 5 correct Explanation: A function f will be continuous at x = 5 when f (5) exists and lim x 5 f ( x ) = f (5) . Now f (5) exists for all the functions defined above; in addition, inspection shows that all these functions have the property lim x 5 f ( x ) = f (5) except for f ( x ) = braceleftBigg 1 x 5 x negationslash = 5 5 x = 5 . . garcia (jjg2564) Hw 04 gualdani (56410) 3 Consequently, this function is the only one that is not continuous at x = 5....
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HW 5 Calc - garcia (jjg2564) Hw 04 gualdani (56410) 1 This...

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