solns1_09

# solns1_09 - Math 135 Professor Carlson Test 1 Solutions 1....

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Unformatted text preview: Math 135 Professor Carlson Test 1 Solutions 1. (12 pts) Find the domain and sketch the graph of the function ( a ) f ( x ) = x 3 + 5 x 2 + x − 2 . The denominator factors as x 2 + x − 2 = ( x + 2)( x − 1), so the function is defined unless x = − 2 or x = 1. ( b ) g ( x ) = radicalbig x − x 2 . The square root is defined for nonnegative numbers. Thus the function is defined where x − x 2 = x (1 − x ) ≥ 0, that is for 0 ≤ x ≤ 1. 2. (14 pts) (a) Suppose p ( x ) is a polynomial of degree three, with p ( − 1) = p (0) = p (2) = 0 , p (1) = 6 . Find p ( x ) . The polynomial has factors corresponding to the three roots at x = − 1 , , 2, so p ( x ) = C ( x + 1) x ( x − 2). Since p (1) = 6 we have 6 = C (2)( − 1) or C = − 3. Thus p ( x ) = − 3( x + 1) x ( x − 2) = − 3 x 3 + 3 x 2 + 6 x. You could also assume p ( x ) = Ax 3 + Bx 2 + Cx + D and solve the equations determined by plugging in the appropriate values at x = − 1 , , 1 , 2, but this is more complicated.2, but this is more complicated....
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## This note was uploaded on 06/21/2011 for the course MATHEMATIC 135 taught by Professor Phystago during the Spring '11 term at Waterloo.

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solns1_09 - Math 135 Professor Carlson Test 1 Solutions 1....

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