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110quiz3solutions

# 110quiz3solutions - Math 110 Quiz 3 Sept 21 2006 Name 1 Let...

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Math 110 Quiz 3 Sept. 21, 2006 Name: 1. Let f ( x ) = x 2 - 5 and g ( x ) = 2 x + 6. (a) Determine the functions ( f + g )( x ), ( f - g )( x ), ( fg )( x ) and ( f g )( x ). (b) Determine ( f + g )( - 3), ( f - g )( - 3), ( fg )( - 3) and f g ( - 3). (c) Determine the domain of the functions f + g , f - g , fg , and f g . Solutions: (a) ( f + g )( x ) = f ( x ) + g ( x ) = x 2 - 5 + 2 x + 6 , ( f - g )( x ) = f ( x ) - g ( x ) = x 2 - 5 - 2 x + 6 , ( fg )( x ) = f ( x ) g ( x ) = ( x 2 - 5) 2 x + 6 , and f g ( x ) = f ( x ) g ( x ) = x 2 - 5 2 x + 6 (b) ( f + g )( - 3) = ( - 3) 2 - 5 + 2( - 3) + 6 = 4 , ( f - g )( - 3) = ( - 3) 2 - 5 - 2( - 3) + 6 = 4 , ( fg )( - 3) = (( - 3) 2 - 5)0 = 0 , and f g ( - 3) = 4 0 = undefined (c) To find the domain of f + g , f - g , and fg , we find the domains of f and g and take what is com- mon to both. Since the domain of f is ( -∞ , ), we only need to find the domain of g . Since we cannot take the square root of a negative number, we set 2 x + 6 0, which means x ≥ - 3. So the domain of f + g , f - g , and fg is [ - 3 , ) . For the domain of f g , we take the domain we just found for the other functions and throw out values that make g zero. This means we throw out - 3. Hence, the domain of f g is ( - 3 , ) .

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