08exam1sol_v1

# 08exam1sol_v1 - Math 20A. Midterm Exam 1 October 23, 2008...

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Math 20A. Midterm Exam 1 October 23, 2008 1. Let ( ) 2 2 f xx ! " # (a) (2 points) Determine the domain and range of f . The domain is x ! 2 and the range is y " 2. A rough sketch appears below: # 4 # 3 # 2 # 1 1 2 1 2 3 4 (b) (4 points) Find a formula for the inverse 1 () f x # and state its domain and range. \$% 2 2 2 2 2 2 2 2 2 2 2 2 y x x y x y x y yx ! " # & ! " # & # ! # & # ! # & ! # # So, , but only for x " 2. \$ 2 1 ( ) 2 2 f x x # ! # # % We have a restricted domain, since 2 – ( x – 2) 2 is not one-to-one (and hence not invertible) on (- \$ , \$ ). Recall, the domain and range of a function are the range and domain of the inverse function. Hence, we have that the domain of 1 f x # is x " 2 and the range of 1 f x # is y ! 2. 1

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2. Let 18 () 1 t ft e ! " . (a) (3 points) Find f t . Using the quotient rule, we have that: 22 (1 )(0) 18( ) 18 ) ) tt ee " # # !! "" t e . (b) (3 points) Compute (ln(2)) f . The correct value is an integer. ln(2) ln(2) 2 2 18 18(2) (ln(2)) 4 ) (1 2) e f e ## !# 3. Find the following limits: (a) (3 points) 0 42 lim h h h ( "# \$% 00 0 0 0 4 2 4 2 4 2 lim lim 44 lim lim 1 lim 1 4 hh h h h h h h h (( ( ( ( )* " # " # " " !
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## This note was uploaded on 11/09/2010 for the course MATH 20A 20A taught by Professor Yacobi during the Fall '08 term at UCSD.

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08exam1sol_v1 - Math 20A. Midterm Exam 1 October 23, 2008...

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