{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

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 ) ,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 . .
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 ) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}