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Unformatted text preview: Name: PID: Midterm 2, Math 109  Winter 2008 Duration: 50 minutes Please close your books, turn off your calculators and phones. To get full credit you should explain your answers. 1. Let f : R → R f ( x ) = x 2 and g : R → R , g ( x ) = 7 x − 4 be two functions. a.(2 points) Find f ◦ g and g ◦ f . Proof. We have ( f ◦ g )( x ) = f ( g ( x )) = f (7 x − 4) = (7 x − 4) 2 = 49 x 2 − 56 x + 8 and ( g ◦ f )( x ) = g ( f ( x )) = g ( x 2 ) = 7 x 2 − 4 b.(3 points) Show that g is a bijective function and find its inverse g 1 . Show that f is not a bijective function. Proof. g is injective because if g ( x 1 ) = g ( x 2 ), this means that 7 x 1 − 4 = 7 x 2 − 4. It follows that 7 x 1 = 7 x 2 which implies that x 1 = x 2 . g is surjective because for any y ∈ R , if g ( x ) = y , then 7 x − 4 = y which implies that x = y +4 7 . Thus, g is bijective and its inverse is g 1 ( y ) = y +4 7 ....
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 Winter '06
 Knutson
 Math, 50 Minutes, CORE COURSES, inclusionexclusion

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