Exam 1A Solutions

Exam 1A Solutions - Mt100 - Calculus I October 3, 2007...

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Unformatted text preview: Mt100 - Calculus I October 3, 2007 Solutions Exam 1 Variation: A 1.(10) 3 5 a. y = " x + 2 2 ! b. y = 2 3 x 2.(10) ! a. 3" 4 b. Solving y = f (x ) = ! ( ) ( ) ! , so the domain of the inverse is x "1 x "1 "#,1 $ 1, # = R " 1 and the range is the same, the domain of f. x +1 for x gives f "1 (x ) = x +1 {} ! 1 2 3 2 ! 3.(10) a. "2 + =" b. 2 + x 2 = "3x # x = "1, "2 ! ! 4.(10) a. 48 + 1 " 8 + 1 48 " 8 = 1 10 b. Greater: ! 7 6 5 4 3 2 10 20 30 40 50 60 5.(5) Since lim f (x ) = lim 7 cos x = 7 cos # = $7 and lim f (x ) = lim x 2 $ 2#x + # 2 $ 7 = $7 , the two sided limit of f (x ) at x = " exists, so the discontinuity is removable. x "# + x "# + x "# $ x "# $ ! ! 6.(15) a. x "5 lim x 3 # 125 x #5 2x sin3x ! (x # 5)$&%x = lim x "5 ! 2 ' + 5x + 25) ( x #5 $ ' = lim &x 2 + 5x + 25) = 75 ( x "5% ! b. x "0 lim = lim x "0 3 sin3x 2 = 2 3x 3 ! c. ! #1 & #1 & 0 " sin x cos% ( " sin x ) * sin x " sin x cos% ( " sin x , so since % ( % ( $x ' $x ' #1 & $ ' lim &# sin x ) = lim sin x = 0 , lim sin x cos% ( = 0 % ( ( x "0 x "0% x "0 $x ' ! ! ...
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This note was uploaded on 06/04/2008 for the course MT 100 taught by Professor Keane during the Fall '07 term at BC.

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Exam 1A Solutions - Mt100 - Calculus I October 3, 2007...

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