K Exam 1-solutions

# K Exam 1-solutions - Version 030 – K Exam 1 – ben-zvi...

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Unformatted text preview: Version 030 – K Exam 1 – ben-zvi – (55600) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If a ball is thrown into the air with a velocity of 49 ft/sec, its height in feet after t seconds is given by y = 49 t − 16 t 2 . Find the average velocity of the ball for the time period beginning at t = 1 and lasting 1 / 2 seconds. 1. average vel. = 5 ft/sec 2. average vel. = 8 ft/sec 3. average vel. = 6 ft/sec 4. average vel. = 9 ft/sec correct 5. average vel. = 7 ft/sec Explanation: If the height of the ball after t seconds is given by y = y ( t ) = 49 t − 16 t 2 , then the average velocity of the ball over a time interval [1 , t + h ] is avg. vel. = y (1 + h ) − y (1) h . Now y (1 + h ) = 49(1 + h ) − 16(1 + h ) 2 = 49(1 + h ) − 16(1 + 2 h + h 2 ) = 33 + 17 h − 16 h 2 , while y (1) = 49 − 16 = 33 . Thus y (1 + h ) − y (1) h = 17 h − 16 h 2 h = 17 − 16 h. Consequently, when h = 1 / 2, therefore, average velocity = 9 ft/sec . 002 10.0 points Below is the graph of a function f . 2 4 − 2 − 4 2 4 − 2 − 4 Use the graph to determine lim x → 4 f ( x ). 1. limit = 1 2. limit = 0 3. limit = 2 4. does not exist 5. limit = − 1 correct Explanation: From the graph it is clear that the limit lim x → 4 − f ( x ) = − 1 , from the left and the limit lim x → 4+ f ( x ) = − 1 , from the right exist and coincide in value. Thus the two-sided limit exists and lim x → 4 f ( x ) = − 1 . 003 10.0 points Version 030 – K Exam 1 – ben-zvi – (55600) 2 Determine lim x → x − 1 x 2 ( x + 7) . 1. limit = − 1 7 2. limit = 0 3. limit = 1 4. limit = −∞ correct 5. limit = ∞ 6. none of the other answers Explanation: Now lim x → x − 1 = − 1 . On the other hand, x 2 ( x + 7) > 0 for all small x , both positive and negative, while lim x → x 2 ( x + 7) = 0 . Consequently, limit = −∞ . keywords: evaluate limit, rational function 004 10.0 points Determine lim x → 8 √ x + 1 − 3 x − 8 . 1. limit = 6 2. limit = 1 3 3. limit doesn’t exist 4. limit = 1 6 correct 5. limit = 3 Explanation: After rationalizing the numerator we see that √ x + 1 − 3 = ( x + 1) − 9 √ x + 1 + 3 = x − 8 √ x + 1 + 3 . Thus √ x + 1 − 3 x − 8 = 1 √ x + 1 + 3 for all x negationslash = 8. Consequently, limit = lim x → 8 1 √ x + 1 + 3 = 1 6 . 005 10.0 points Evaluate lim x →− 2 x + 2 x 2 − 3 x − 10 . 1. limit = − 3 2. limit does not exist 3. limit = − 2 7 4. limit = 1 7 5. limit = − 1 7 correct 6. limit = 2 7 Explanation: Since x 2 − 3 x − 10 = ( x − 5)( x + 2) , the expression above can be rewritten as x + 2 x 2 − 3 x − 10 = 1 x − 5 Version 030 – K Exam 1 – ben-zvi – (55600)...
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K Exam 1-solutions - Version 030 – K Exam 1 – ben-zvi...

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