Review 2 Solutions

# Review 2 Solutions - 1 Review Sheet Solutions Math 112 Fall...

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1 Review Sheet Solutions Math 112 Fall 2008 Meghan De Witt 1. Determine the domain of each function: To determine the domain of a fraction, set the bottom to be not equal to zero. To determine the domain of a square-root, set the inside to be greater than or equal to zero. a) y = 1 - 5 x +1 - 5 x + 1 6 = 0 - 5 x 6 = - 1 x 6 = 1 5 Thus the domain is ± -∞ , 1 5 ² ± 1 5 , ² b) y = x - 9 x - 9 0 x 9 Thus the domain is [9 , ) c) y = q x - 2 2 x +6 We do a sign chart. x - 2 2 x + 6 y ( -∞ , - 3) - - + ( - 3 , 2) - + - (2 , ) + + + Also, we can’t have the denominator equal to zero. Thus the domain is ( -∞ , - 3) [2 , )

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2 2. Find the domain and range of each function: To ﬁnd the range of a function, solve for x and ﬁnd the domain of that function. a) y = 4 x 3 - 5 The domain is R since there is no division or square-roots. To ﬁnd the range: y = 4 x 3 - 5 4 x 3 = y + 5 x 3 = y + 5 4 x = 3 r y + 5 4 This has a domain of R , hence that is the range of the original function. b) y = 2 x +7 - 3 x +2 To ﬁnd the domain: - 3 x + 2 6 = 0 - 3 x 6 = - 2 x 6 = 2 3 Hence the domain is ± -∞ , 2 3 ² ± 2 3 , ² To ﬁnd the range: y = 2 x + 7 - 3 x + 2 - 3 xy + 2 y = 2 x + 7 - 3 xy - 2 x = 7 - 2 y x ( - 3 y - 2) = 7 - 2 y x = 7 - 2 y - 3 y - 2
3 To ﬁnd the domain of this function, - 3 y - 2 6 = 0 - 3 y 6 = 2 y 6 = - 2 3 Thus the range of the original function is ± -∞ , - 2 3 ² ± - 2 3 , ² 3. For f ( x ) = x 3 - x 2 + x - 1 ﬁnd the following: a) f (2) = 2 3 - 2 2 + 2 - 1 = 5 b) f ( - x ) = ( - x ) 2 - ( - x ) 2 + ( - x ) - 1 = - x 2 - x 2 - x - 1 Don’t forget the parenthesis! c) f ( x + 1) = ( x + 1) 3 - ( x + 1) 2 + ( x + 1) - 1 = x 3 + 3 x 2 + 3 x + 1 - x 2 - 2 x - 1 + x + 1 - 1 = x 3 + 2 x 2 + 2 x d) f (1 /x ) = (1 /x ) 3 - (1 /x ) 2 + (1 /x ) - 1 = 1 - x + x 2 - x 3 x 3 4. Graph the function y = x 2 - 2 x < - 1 | x | - 1 x < 3 x 3 x 5. Find the average value of f ( x ) = 2 x 2 - 6 on the interval [1 , 4].

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4 Thus we are basically taking a slope. f (4) - f (1) 4 - 1 = (2(4) 2 - 6) - (2(1) 2 - 6) 3 = (32 - 6) - (2 - 6) 3 = 26 + 4 3 = 30 3 = 10 6. Find the average value of f ( x ) = 1 - x +4 on the interval [ - 1 , 3]. f ( b ) - f ( a ) b - a = 1 / p (1) - 1 / p (5) 3 + 1 = 1 4 ± 1 - 1 p (5) ! 7. For the function f ( x ) = 2 x x 2 - 3 ﬁnd and simplify the following: a) f ( x ) - f ( a ) x - a = 2 x x 2 - 3 - 2 a a 2 - 3 x - a = 2 xa 2 - 6 x - 2 x 2 a +6 a ( x 2 - 3)( a 2 - 3) x - a = - 2 xa ( x - a ) - 6( x - a ) ( x 2 - 3)( a 2 - 3)( x - a ) = - 2 xa - 6 ( x 2 - 3)( a 2 - 3)
5 b) f ( x + h ) - f ( x ) h = 2( x + h ) ( x + h ) 2 - 3 - 2 x x 2 - 3 h = 1 h · (2 x + 2 h )( x 2 - 3) - 2 x ( x 2 + 2 xh + h 2 - 3) ( x 2 - 3)(( x + h ) 2 - 3) = 2 x 3 - 6 x + 2 x 2 h - 6 h - 2 x 3 - 4 x 2

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## This note was uploaded on 04/29/2009 for the course MATH 112 taught by Professor Carlson during the Fall '08 term at University of Wisconsin.

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Review 2 Solutions - 1 Review Sheet Solutions Math 112 Fall...

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