Homework #1 - jiwani(amj566 – Homework01 – Fouli...

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Unformatted text preview: jiwani (amj566) – Homework01 – Fouli – (58395) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rationalize the numerator of √ x + 4 − √ x − 3 x . 1. x √ x + 4 + √ x − 3 2. 7 x ( √ x + 4 + √ x − 3) correct 3. 7 x ( √ x + 4 − √ x − 3) 4. 7 x √ x + 4 − √ x − 3 5. 1 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) . 002 10.0 points Which, if any, of the following statements are true when a, b are real numbers? A. For all positive a and b , √ a + √ b = radicalBig a + 2 √ ab + b . B. For all positive a and b , a − b √ a + √ b = √ a − √ b. C. For all a and b , radicalBig ( a + b ) 2 = | a + b | . 1. all of them correct 2. B and C only 3. C only 4. A and C only 5. A and B only 6. A only 7. B only 8. none of them Explanation: A. TRUE: by the known product, ( x + y ) 2 = x 2 + 2 xy + y 2 . On the other hand, radicalBig ( x + y ) 2 = | x + y | , so if x + y > 0, x + y = radicalbig x 2 + 2 xy + y 2 . But if a, b are positive we can set x = √ a and y = √ b . The result follows since x and y are then positive. B. TRUE: by the known difference of squares factorizaation, x 2 − y 2 = ( x − y )( x + y ) . But if a, b are positive we can set x = √ a and y = √ b . The result follows after division. jiwani (amj566) – Homework01 – Fouli – (58395) 2 C. TRUE: we know that radicalBig ( x + y ) 2 = | x + y | , and since radicalbig ( · ) is always non-negative, the right hand side has to be non-negative. That’s why the absolute value sign is needed. keywords: 003 10.0 points Find the domain of the function f ( x ) = √ x + 3 x − 6 . 1. domain = { x : x > − 3 , x negationslash = 6 } 2. domain = { x : x ≥ 3 } 3. domain = { x : x ≤ 3 } 4. domain = { x : x < 3 , x negationslash = − 6 } 5. domain = { x : x ≤ − 3 , x negationslash = − 6 } 6. domain = { x : x ≥ − 3 , x negationslash = 6 } correct Explanation: The domain of f consists of all x for which √ x + 3 x − 6 is well-defined. This requires that √ x + 3 be defined and that x − 6 negationslash = 0. Now √ x is defined for all x ≥ 0, so √ x + 3 is defined when x + 3 ≥ 0, i.e. , when x ≥ − 3. Consequently, f has domain = { x : x ≥ − 3 , x negationslash = 6 } . keywords: natural domain, square root func- tion, quotient of functions, sign chart 004 10.0 points Simplify the expression f ( x ) = 3 + 6 x − 3 4 + 32 parenleftBig x x 2 − 9 parenrightBig as much as possible....
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Homework #1 - jiwani(amj566 – Homework01 – Fouli...

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