# HW1Answers - Villarreal Natalie – Homework 1 – Due Sep...

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Unformatted text preview: Villarreal, Natalie – Homework 1 – Due: Sep 7 2007, 3:00 am – Inst: Louiza Fouli 1 This print-out should have 21 questions. Multiple-choice 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 + 5- √ x- 3 x . 1. 8 x ( √ x + 5- √ x- 3) 2. 8 x ( √ x + 5 + √ x- 3) correct 3. 2 x √ x + 5 + √ x- 3 4. 2 x ( √ x + 5 + √ x- 3) 5. 8 x √ x + 5- √ x- 3 Explanation: By the difference of squares, ( √ x + 5- √ x- 3)( √ x + 5 + √ x- 3) = ( √ x + 5) 2- ( √ x- 3) 2 = 8 . Thus, after multiplying both the numerator and the denominator in the given expression by √ x + 5 + √ x- 3 , we obtain 8 x ( √ x + 5 + √ x- 3) . keywords: rationalization numerator, 002 (part 1 of 1) 10 points Simplify the expression f ( x ) = 4 + 4 x- 2 3 + 9 ‡ x x 2- 4 · as much as possible. 1. f ( x ) = x- 2 2 x- 4 2. f ( x ) = 4 3 ‡ x- 2 x + 4 · 3. f ( x ) = 4 3 ‡ x + 2 x + 4 · correct 4. f ( x ) = x- 2 x- 4 5. f ( x ) = x + 2 x- 4 6. f ( x ) = 4 3 ‡ x + 2 2 x + 4 · Explanation: After bringing the numerator to a common denominator it becomes 4 x- 8 + 4 x- 2 = 4 x- 4 x- 2 . Similarly, after bringing the denominator to a common denominator and factoring it be- comes 3 x 2- 12 + 9 x x 2- 4 = 3( x- 1)( x + 4) x 2- 4 . Consequently, f ( x ) = 4 + 4 x- 2 3 + 9 ‡ x x 2- 4 · = 4 x- 4 3( 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 ) = 4 3 µ x + 2 x + 4 ¶ . Villarreal, Natalie – Homework 1 – Due: Sep 7 2007, 3:00 am – Inst: Louiza Fouli 2 keywords: 003 (part 1 of 1) 10 points Find the solution set of the inequality x- 2 x- 6 < x + 3 x + 2 . 1. ‡- 14 3 ,- 2 · [ ‡ 6 , ∞ · 2. ‡-∞ ,- 14 3 i [ ‡- 2 , 6 · 3. ‡-∞ ,- 14 3 · [ ‡- 2 , 6 · correct 4. h- 14 3 ,- 2 · [ ‡ 6 , ∞ · 5. ‡-∞ ,- 14 3 · [ h- 2 , 6 · Explanation: To begin we need to arrange that all the terms are on one side of the inequality. Thus the inequality becomes x- 2 x- 6- x + 3 x + 2 < , which in turn becomes ( ‡ ) 3 x + 14 ( x + 2)( x- 6) < 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 =- 14 3 , 6 ,- 2, and the sign chart- 14 3- 2 6- +- + determines which sign it takes in a given in- terval. Thus the solution set of ( ‡ ) is the union ‡-∞ ,- 14 3 · [ ‡- 2 , 6 · . keywords: 004 (part 1 of 1) 10 points The straight line ‘ is parallel to y + 3 x = 1 and passes through the point P (4 , 2). Find its x-intercept....
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HW1Answers - Villarreal Natalie – Homework 1 – Due Sep...

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