This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 028 – Homework 1  PreCalc Review – Helleloid – (58250) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Welcome to Quest. This homework over laps #2 and #3. Use it as a precalculus review and diagnostic. Good luck with M408K. 001 10.0 points Rewrite the expression radicalBig a + √ b. 1. ( a + b 1 / 2 ) 1 / 2 correct 2. a 1 / 2 + b 1 / 4 3. a + b 4. a + b 1 / 2 5. a + b 1 / 4 Explanation: Since √ x = x 1 / 2 , we see that radicalBig a + √ b = ( a + b 1 / 2 ) 1 / 2 . 002 10.0 points Rationalize the numerator of √ x + 4 − √ x − 3 x . 1. 7 x ( √ x + 4 + √ x − 3) correct 2. 1 x ( √ x + 4 + √ x − 3) 3. x √ x + 4 + √ x − 3 4. 7 x ( √ x + 4 − √ x − 3) 5. 7 x √ x + 4 − √ x − 3 Explanation: By the difference of squares, ( √ x + 4 − √ x − 3)( √ x + 4 + √ x − 3) = ( √ x + 4) 2 − ( √ x − 3) 2 = 7 . Thus, after multiplying both the numerator and the denominator in the given expression by √ x + 4 + √ x − 3 , we obtain 7 x ( √ x + 4 + √ x − 3) . 003 10.0 points Simplify the expression parenleftBig xy − 3 √ z parenrightBig 6 ÷ parenleftBig y 2 x 1 / 3 z − 3 parenrightBig 12 as much as possible, leaving no negative ex ponents and no radicals. 1. x 10 y 42 z 39 correct 2. x 2 y 6 z 33 3. x 10 y 6 z 33 4. x 10 y 42 z 39 5. x 10 z 39 y 42 Explanation: By the Laws of Exponents parenleftBig xy − 3 √ z parenrightBig 6 = x 6 y 18 z 3 , while parenleftBig y 2 x 1 / 3 z − 3 parenrightBig 12 = y 24 z 36 x 4 . Version 028 – Homework 1  PreCalc Review – Helleloid – (58250) 2 Consequently, the given expression can be rewritten as x 6 y 18 z 3 × x 4 y 24 z 36 = x 10 y 42 z 39 . 004 10.0 points Simplify the expression f ( x ) = 5 + 15 x − 4 4 + 60 parenleftBig x x 2 − 16 parenrightBig as much as possible. 1. f ( x ) = x − 4 x − 16 2. f ( x ) = 5 4 parenleftBig x + 4 x + 16 parenrightBig correct 3. f ( x ) = 5 4 parenleftBig x + 4 2 x + 16 parenrightBig 4. f ( x ) = x + 4 x − 16 5. f ( x ) = 5 4 parenleftBig x − 4 x + 16 parenrightBig Explanation: After bringing the numerator to a common denominator it becomes 5 x − 20 + 15 x − 4 = 5 x − 5 x − 4 . Similarly, after bringing the denominator to a common denominator and factoring it be comes 4 x 2 − 64 + 60 x x 2 − 16 = 4( x − 1)( x + 16) x 2 − 16 . Consequently, f ( x ) = 5 + 15 x − 4 4 + 60 parenleftBig x x 2 − 16 parenrightBig = 5 x − 5 4( x − 1)( x + 16) parenleftBig x 2 − 16 x − 4 parenrightBig . On the other hand, x 2 − 16 = ( x + 4)( x − 4) . Thus, finally, we see that f ( x ) = 5 4 parenleftbigg x + 4 x + 16 parenrightbigg ....
View
Full
Document
This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz
 Calculus

Click to edit the document details