16a_practice_midterm_i_solutions

# 16a_practice_midterm_i_solutions - =" 1 2 = 1 so lim...

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16A Practice Midterm I Solutions 1. 3 x 2 + 3 y 2 ! 9 x + 6 y ! 9/4 = 0 x 2 ! 3 x + y 2 + 2 y = 3 / 4 x 2 ! 3 x + 9 4 + y 2 + 2 y + 1 = 3 4 + 9 4 + 1 x ! 3 2 " # \$ % 2 + y + 1 ( ) 2 = 4 The center is (3/2 , -1) and the radius is 2. 2. f ( x ) = x 2 ! 1 ( x ! 1) 2 ( x + 2) = x ! 1 ( ) x + 1 ( ) x ! 1 ( ) 2 x + 2 ( ) = x + 1 x ! 1 ( ) x + 2 ( ) a) This function has vertical asymptotes at x = 1 & x = -2 lim x ! 1 + f ( x ) = " lim x ! 1 # f ( x ) = #" lim x !# 2 + f ( x ) = " lim x !# 2 # f ( x ) = #" b) f has a horizontal asymptote at y = 0 lim x !" x 3 1 x # 1 x 3 ( ) x 3 1 # 2 x # 3 x 2 + 2 x 3 \$ % ( ) = 0 3. a) g h x ( ) ( ) = 1 1 x + 1 ! x x = x 1 + x b) g x ! 1 ( ) = 1 x h 1 x ( ) = 1 1 x = x c) x cannot = -1 since that would make g undefined. Since 0 is not in the range of g we don’t have to worry about h being undefined. 4. a) lim x ! 1 x + 2 x 2 + 3 = 3 4 b) lim x ! 2 x " 2 ( ) x " 3 ( ) 2 x " 2 ( ) x + 1 ( ) = lim x ! 2 x " 3 2 x + 1 ( ) = " 1 6 c) lim x ! 4 5 + x " 3 x " 4 # 5 + x + 3 5 + x + 3 = lim x ! 4 5 + x " 9 x " 4 ( ) 5 + x + 3 ( ) = lim x ! 4 1 5 + x + 3 = 1 6 d) lim x !" x 2 1 # 5 x + 6 x 2 ( ) x 2 3 # 7 x 2 ( ) = 1 3 5. ii and vi

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6. I can’t sketch the graph here. b) lim x ! 1 + f ( x ) = 1 " 2 = 1 lim x ! 1 " f ( x ) = 1 2 = 1 c) f is continuous at x = 1 since the limit is 1 (from b) and f(1) =1 (it exists) the function is not continuous at x = -1 because lim x !" 1 " f ( x ) = " 1 + 1 = 0 and lim x !" 1 + f ( x
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Unformatted text preview: ) = ( " 1) 2 = 1 so lim x !" 1 f ( x ) dne So the function is continuous on (-∞ ,-1) (-1, ∞ ) 7. ! f x ( ) = lim ! x " 1 3 x + ! x ( ) # 1 # 1 3 x # 1 ! x = lim ! x " 3 x # 1 # 3 x # 3 \$ x + 1 3 x + \$ x ( ) # 1 % & ’ ( 3 x # 1 ( ) \$ x lim \$ x " # 3 \$ x 3 x + \$ x ( ) # 1 % & ’ ( 3 x # 1 ( ) ) 1 \$ x lim \$ x " # 3 3 x + \$ x ( ) # 1 % & ’ ( 3 x # 1 ( ) = # 3 3 x # 1 ( ) 2 8. lim x ! " ( 9 x 2 # 4 x # 3 x ) \$ 9 x 2 # 4 x + 3 x 9 x 2 # 4 x + 3 x % & ’ ’ ( ) * * = lim x ! " 9 x 2 # 4 x # 9 x 2 9 x 2 # 4 x + 3 x = lim x ! " # 4 x x 2 9 # 4 x ( ) + 3 x = lim x ! " # 4 x x 9 # 4 x + 3 x = lim x ! " # 4 x x 9 # 4 x + 3 % & ’ ( ) * = # 2 3...
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16a_practice_midterm_i_solutions - =" 1 2 = 1 so lim...

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