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Unformatted text preview: Tran, Tony – Homework 1 – Due: Sep 7 2007, 3:00 am – Inst: Samuels 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Yes, Homework 1 is due AFTER Homework 2. 001 (part 1 of 1) 10 points Rationalize the numerator of √ x + 4 √ x 1 x . 1. 3 x ( √ x + 4 + √ x 1) 2. 5 x ( √ x + 4 + √ x 1) correct 3. 3 x √ x + 4 + √ x 1 4. 5 x ( √ x + 4 √ x 1) 5. 5 x √ x + 4 √ x 1 Explanation: By the difference of squares, ( √ x + 4 √ x 1)( √ x + 4 + √ x 1) = ( √ x + 4) 2 ( √ x 1) 2 = 5 . Thus, after multiplying both the numerator and the denominator in the given expression by √ x + 4 + √ x 1 , we obtain 5 x ( √ x + 4 + √ x 1) . keywords: rationalization numerator, 002 (part 1 of 1) 10 points Simplify the expression f ( x ) = 1 + 4 x 5 3 + 72 ‡ x x 2 25 · as much as possible. 1. f ( x ) = 1 3 ‡ x 5 x + 25 · 2. f ( x ) = 1 3 ‡ x + 5 x + 25 · correct 3. f ( x ) = x 5 2 x 25 4. f ( x ) = x 5 x 25 5. f ( x ) = x + 5 x 25 6. f ( x ) = 1 3 ‡ x + 5 2 x + 25 · Explanation: After bringing the numerator to a common denominator it becomes x 5 + 4 x 5 = x 1 x 5 . Similarly, after bringing the denominator to a common denominator and factoring it be comes 3 x 2 75 + 72 x x 2 25 = 3( x 1)( x + 25) x 2 25 . Consequently, f ( x ) = 1 + 4 x 5 3 + 72 ‡ x x 2 25 · = x 1 3( x 1)( x + 25) ‡ x 2 25 x 5 · . On the other hand, x 2 25 = ( x + 5)( x 5) . Thus, finally, we see that f ( x ) = 1 3 µ x + 5 x + 25 ¶ . Tran, Tony – Homework 1 – Due: Sep 7 2007, 3:00 am – Inst: Samuels 2 keywords: 003 (part 1 of 1) 10 points Find the solution set of the inequality x 1 x 5 < x + 3 x + 1 . 1. ‡∞ , 7 · [ h 1 , 5 · 2. ‡∞ , 7 · [ ‡ 1 , 5 · correct 3. h 7 , 1 · [ ‡ 5 , ∞ · 4. ‡ 7 , 1 · [ ‡ 5 , ∞ · 5. ‡∞ , 7 i [ ‡ 1 , 5 · Explanation: To begin we need to arrange that all the terms are on one side of the inequality. Thus the inequality becomes x 1 x 5 x + 3 x + 1 < , which in turn becomes ( ‡ ) 2 x + 14 ( x + 1)( x 5) < after the right hand side is brought to a com mon denominator. Now the right hand side changes sign at the zeros of its numerator and denominator, i.e. , at x = 7 , 5 , 1, and the sign chart 7 1 5 + + determines which sign it takes in a given in terval. Thus the solution set of ( ‡ ) is the union ‡∞ , 7 · [ ‡ 1 , 5 · . keywords: 004 (part 1 of 1) 10 points The straight line ‘ is parallel to y + 3 x = 2 and passes through the point P (5 , 3). Find its xintercept. 1. xintercept = 6 2. xintercept = 9 2 3. xintercept = 4 4. xintercept = 19 3 5. xintercept = 6 correct Explanation: Since ‘ is parallel to the line y + 3 x = 2, these lines have the same slope 3, Thus by the pointslope formula the equation of ‘ is given by y 3 = 3( x 5) ....
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This note was uploaded on 04/27/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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