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Unformatted text preview: Garcia, Ilse – Homework 1 – Due: Sep 7 2007, 3:00 am – Inst: Fonken 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. 5 x ( √ x + 4 √ x 1) 2. 5 x √ x + 4 √ x 1 3. 3 x √ x + 4 + √ x 1 4. 3 x ( √ x + 4 + √ x 1) 5. 5 x ( √ x + 4 + √ x 1) correct 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 ) = 5 + 15 x 2 2 6 ‡ x x 2 4 · as much as possible. 1. f ( x ) = x + 2 x + 4 2. f ( x ) = 5 2 ‡ x 2 x 4 · 3. f ( x ) = 5 2 ‡ x + 2 x 4 · correct 4. f ( x ) = 5 2 ‡ x + 2 2 x 4 · 5. f ( x ) = x 2 x + 4 6. f ( x ) = x 2 2 x + 4 Explanation: After bringing the numerator to a common denominator it becomes 5 x 10 + 15 x 2 = 5 x + 5 x 2 . Similarly, after bringing the denominator to a common denominator and factoring it be comes 2 x 2 8 6 x x 2 4 = 2( x + 1)( x 4) x 2 4 . Consequently, f ( x ) = 5 + 15 x 2 2 6 ‡ x x 2 4 · = 5 x + 5 2( x + 1)( x 4) ‡ x 2 4 x 2 · . On the other hand, x 2 4 = ( x + 2)( x 2) . Thus, finally, we see that f ( x ) = 5 2 µ x + 2 x 4 ¶ . Garcia, Ilse – Homework 1 – Due: Sep 7 2007, 3:00 am – Inst: Fonken 2 keywords: 003 (part 1 of 1) 10 points Find the solution set of the inequality x 3 x 5 < x + 6 x + 5 . 1. ‡∞ , 15 i [ ‡ 5 , 5 · 2. ‡ 15 , 5 · [ ‡ 5 , ∞ · 3. ‡∞ , 15 · [ h 5 , 5 · 4. h 15 , 5 · [ ‡ 5 , ∞ · 5. ‡∞ , 15 · [ ‡ 5 , 5 · correct Explanation: To begin we need to arrange that all the terms are on one side of the inequality. Thus the inequality becomes x 3 x 5 x + 6 x + 5 < , which in turn becomes ( ‡ ) x + 15 ( x + 5)( 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 = 15 , 5 , 5, and the sign chart 15 5 5 + + determines which sign it takes in a given in terval. Thus the solution set of ( ‡ ) is the union ‡∞ , 15 · [ ‡ 5 , 5 · . keywords: 004 (part 1 of 1) 10 points The straight line ‘ is parallel to y + 5 x = 3 and passes through the point P (5 , 4). Find its xintercept. 1. xintercept = 29 5 correct 2. xintercept = 21 5 3. xintercept = 29 6 4. xintercept = 29 5 5. xintercept = 6 Explanation: Since ‘ is parallel to the line y + 5 x = 3, these lines have the same slope 5, Thus by the pointslope formula the equation of ‘ is given by y 4 = 5( x 5) ....
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This note was uploaded on 02/22/2009 for the course M 58365 taught by Professor Gilbert during the Spring '08 term at University of Texas.
 Spring '08
 Gilbert
 Calculus

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