K Exam 1-solutions

# K Exam 1-solutions - Version 076 K Exam 1 Hamrick(54670...

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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 10.0points 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 10.0points 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 10.0points 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 .

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Version 076 – K Exam 1 – Hamrick – (54670) 2 004 10.0points 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 10.0points 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 10.0points Determine lim x 2 x 2 x + 2 2 . 1. limit = 1 4 2. limit = 2 3. limit = 4 correct 4. limit = 1 2 5. limit doesn’t exist Explanation: After rationalizing the denominator we see that 1 x + 2 2 = x + 2 + 2 ( x + 2) 4 = x + 2 + 2 x 2 .
Version 076 – K Exam 1 – Hamrick – (54670) 3 Thus x 2 x + 2 2 = x + 2 + 2 for all x negationslash = 2. Consequently, limit = lim x 2 ( x + 2 + 2) = 4 .

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