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Unformatted text preview: f . The solutions are x = 1. To achieve x 21 0, rst observe that x 21 = 0 has solutions x = 1, and that we have already excluded these x s, so we need to ensure x 21 > 0. The easiest way to do this is to factor: x 21 = ( x + 1)( x1) . From this factorization, we see that in order for x 21 > 0, we need x +1 and x1 to have the same signs (i.e. either both should be negative or both should be positive). Both x + 1 and x1 are negative when x <1. Both x + 1 and x1 are positive when x > 1. Therefore, the domain is ( ,1) (1 , ) . 3. Given the functions f ( x ) = 1 2 x +1 and g ( x ) = x , nd ( f g )( x ). SOLUTION There is not much to do here. Simply rewrite f with x replaced by g ( x ) = x . The solution is ( f g )( x ) = f ( g ( x ) ) = 1 p 2 x + 1 ....
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This note was uploaded on 12/27/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.
 Spring '08
 Smith
 Calculus, Slope

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