HW01-solutionsTW - white (taw933) HW01 ben-zvi (55600) This...

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white (taw933) – HW01 – ben-zvi – (55600) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points SimpliFy the expression f ( x )= 3+ 15 x - 4 1 - 15 ± x x 2 - 16 ² as much as possible. 1. f ( x 3( x - 4) x - 16 2. f ( x x - 4 x +16 3. f ( x 3( x +4) 2 x - 16 4. f ( x 3( x x - 16 correct 5. f ( x x +4 x 6. f ( x x - 4 2 x Explanation: AFter bringing the numerator to a common denominator it becomes 3 x - 12 + 15 x - 4 = 3 x +3 x - 4 . Similarly, aFter bringing the denominator to acommondenom ina to randFac r ingi tbe - comes x 2 - 16 - 15 x x 2 - 16 = ( x +1)( x - 16) x 2 - 16 . Consequently, f ( x 15 x - 4 1 - 15 ± x x 2 - 16 ² = 3 x ( x x - 16) ± x 2 - 16 x - 4 ² . On the other hand, x 2 - 16 = ( x +4)( x - 4) . Thus, fnally, we see that f ( x 3( x x - 16 . 002 10.0 points Let f be the quadratic Function defned by f ( x )=2 x 2 +12 x +50 . Use completing the square to fnd h so that f ( x ( x + h ) 2 + k For some value oF k . 1. h =3 correct 2. h = - 6 3. h =6 4. h = - 3 5. h =1 2 Explanation: Completing the square gives f ( x x 2 x =2 ( x 2 +6 x +25) ( x 2 x +9+25 - 9) . Thus f ( x ( x 2 x +9)+32 ( x +3) 2 +32 . Consequently, h .
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white (taw933) – HW01 – ben-zvi – (55600) 2 003 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 = ± a +2 ab + b. B. For all positive a and b , a - b a + b = a - C. For all a and b , ± ( a + b ) 2 = a + 1. BandCon ly 2. Bon 3. Con 4. AandBonly correct 5. AandConly 6. none of them 7. all of them 8. Aonly Explanation: A. TRUE: by the known product, ( x + y ) 2 = x 2 xy + y 2 . On the other hand, ± ( x + y ) 2 = | x + y | , so if x + y> 0, x + y = ² x 2 xy + y 2 . But if a, b are positive we can set x = a and y = b .Th er e su l tf o l l ow ss in c e x and y are then positive. B. TRUE: by the known diFerence of squares factorization, x 2 - y 2 =( x - y )( x + y ) . But if a, b are positive we can set x = a and y = b .Theresu ltfo l lowsa f terd iv is ion . C. ±ALSE: ± ( x + y ) 2 = | x + y | , and since ² ( · )i sa lw ay snon -n ega t e ,th e right hand side has to be non-negative. But if a, b can be positive or negative, an absolute value sign is then needed on the right. keywords: square root, properties of square root, PlaceUT, True±alse, T/±, 004 10.0 points By removing absolute values, express f ( x )= x 2 - 4 | x - 2 | as a piecewise-de²ned function. 1. f ( x ³ x - 2 ,x > - 2 , 2 - x, x < - 2 . 2. f ( x ³ x - 2 > 2 , 2 - 2 . 3. f ( x ³ x > - 2 , - ( x +2) < - 2 . 4. f ( x ³ 2 - x > - 2 , x - 2 < - 2 . 5. f ( x ³ - ( x > 2 , x < 2 . 6. f ( x ³ x > 2 , - ( x < 2 . cor- rect
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white (taw933) – HW01 – ben-zvi – (55600) 3 Explanation: Since | a | = ± a, , a 0 , - a, a< 0 , we see that | x - 2 | = ± x - 2 ,x 2 , - ( x - 2) < 2 . In particular, therefore, x - 2 | x - 2 | = ± 1 > 2 , - 1 < 2 , where the point x =2h a sb e ene x c lud ed from the domain because the quotient is not deFned at that point. On the other hand, x 2 - 4=( x +2)( x - 2) , so f can now be rewritten as a product f ( x )= g ( x ) ² x - 2 | x - 2 | ³ of g ( x x +2andthepreviouspiecewise- deFned function. Consequently, f ( x ± x +2 > 2 , - ( x +2) < 2 . keywords: piecewise-deFned function, abso- lute value, 005 10.0 points Which of the following statements are true?
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HW01-solutionsTW - white (taw933) HW01 ben-zvi (55600) This...

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