K Exam 1-solutions

K Exam 1-solutions - Version 076 – K Exam 1 – Hamrick...

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Unformatted text preview: Version 076 – K Exam 1 – Hamrick – (54670) 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 Use the properties of logarithms to expand log 3 parenleftbigg x 3 y z 5 parenrightbigg . 1. log 3 3 x + log 3 y − log 3 5 z 2. 3 log 3 x + log 3 y − 5 log 3 z correct 3. 3 log 3 x + log 3 y + 5 log 3 z 4. log 3 3 x + log 3 y + log 3 5 z 5. log 3 x + 3 log 3 y − 5 log 3 z Explanation: log 3 parenleftbigg x 3 y z 5 parenrightbigg = log 3 ( x 3 y ) − log 3 z 5 = log 3 x 3 + log 3 y − log 3 z 5 = 3 log 3 x + log 3 y − 5 log 3 z 002 10.0 points Find the value of x when x = 2 3 parenleftBig 6 log 6 3 parenrightBig − 6 parenleftBig 7 − log 7 4 parenrightBig . 1. x = 5 8 2. x = 9 16 3. x = − 9 16 4. x = 1 2 correct 5. x = − 1 2 Explanation: Since 6 log 6 x = x, 7 − log 7 x = 1 x , we see that 6 log 6 3 = = 3 , while 7 − log 7 4 = = 1 4 . Consequently, x = 2 − 3 2 = 1 2 . 003 10.0 points Simplify the expression f ( x ) = cos ( tan − 1 x ) by writing it in algebraic form. 1. f ( x ) = x √ 1 − x 2 2. f ( x ) = radicalbig 1 − x 2 3. f ( x ) = x √ 1 + x 2 4. f ( x ) = radicalbig 1 + x 2 5. f ( x ) = 1 √ 1 + x 2 correct 6. f ( x ) = 1 √ 1 − x 2 Explanation: By definition f ( x ) = cos θ where x = tan θ . But then by Pythagoras, cos θ = 1 √ 1 + x 2 . Consequently, f ( x ) = 1 √ 1 + x 2 . Version 076 – K Exam 1 – Hamrick – (54670) 2 004 10.0 points Determine the inverse, f − 1 , of f when f ( x ) = 6 + 5 x 3 . 1. f − 1 ( x ) = parenleftBig 5 − x 6 parenrightBig 1 / 3 2. f − 1 ( x ) = parenleftBig 6 + x 5 parenrightBig 1 / 3 3. f − 1 ( x ) = parenleftBig 5 + x 6 parenrightBig 1 / 3 4. f − 1 ( x ) = parenleftBig x − 6 5 parenrightBig 1 / 3 correct 5. f − 1 ( x ) = parenleftBig x − 5 6 parenrightBig 1 / 3 6. f − 1 ( x ) = parenleftBig 6 − x 5 parenrightBig 1 / 3 Explanation: To determine f − 1 we solve for x in y = 6 + 5 x 3 and then interchange x, y . In this case 5 x 3 = y − 6 , i.e. x = parenleftBig y − 6 5 parenrightBig 1 / 3 . Consequently, f − 1 ( x ) = parenleftBig x − 6 5 parenrightBig 1 / 3 . 005 10.0 points Below is the graph of a function f . 2 4 6 8 10 − 2 2 4 6 8 − 2 Use the graph to determine lim x → 4 f ( x ). 1. limit = 7 2. limit = 4 3. limit = 5 4. limit does not exist correct 5. limit = 2 Explanation: From the graph it is clear that f has a left hand limit at x = 4, which is equal to 4; and a right hand limit, which is equal to 7. Since the two numbers do not coincide, the limit limit does not exist . 006 10.0 points Determine lim x → 2 x − 2 √ x + 2 − 2 ....
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This note was uploaded on 10/25/2011 for the course M 408 K taught by Professor Jouve during the Fall '08 term at University of Texas.

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K Exam 1-solutions - Version 076 – K Exam 1 – Hamrick...

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