Assignment2solution(0211)

Assignment2solution(0211) - ( x ) 5. f ◦ g ( x ) = x + 2;...

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MATH 0211: Basic Applicable Mathematics Suggested Solution for Assignment 2 1. (a) f ( x ) = 2 x - 3 if x 2 , 1 if 1 x < 2 , 3 - 2 x if x < 1 . (b) See supplementary sheet (c) From the graph we observe that the range is [1 , ) (d) No. 2. (a) f ( x ) = a ( x 2 + b a x ) + c = a ( x + b 2 a ) 2 - a b 2 4 a 2 + c = a ( x + b 2 a ) 2 - b 2 - 4 ac 4 a (b) Let a be a negative number. Since ( x + b 2 a ) 2 is always greater than or equal to 0, then f has a maximum value when a ( x + b 2 a ) 2 = 0, that is, when x + b 2 a = 0, or x = - b 2 a . (c) Since the coefficient of t 2 in h ( t ) is - 1 2 g which is negative, we can apply the result from (b) that h ( t ) reaches the maximum when t = u g . The maximum height is then given by h max = - 1 2 g u 2 g 2 + u 2 g + h 0 = u 2 2 g + h 0 3. ( f + g )( x ) = 1 + x - 2 + x 2 ,x [2 , ) ( f - g )( x ) = 1 + x - 2 - x 2 ,x [2 , ) ( f · g )( x ) = x 2 + x 2 x - 2 ,x [2 , ) ( f/g )( x ) = 1 x 2 + x - 2 x 2 ,x [2 , ) 4. (a) f 1 ( x ) = x - 3 ,f 2 ( x ) = x - 1 ,f ( x ) = f 2 f 1 ( x ) (b) f 1 ( x ) = x 2 + 4 ,f 2 = 1 x + 5 ,f ( x ) = f 2 f 1 ( x ) (c) f 1 ( x ) = x 3 ,f 2 = sin x,f 3 ( x ) = 3 x,f ( x ) = f 3 f 2 f 1
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Unformatted text preview: ( x ) 5. f ◦ g ( x ) = x + 2; x ∈ R g ◦ f ( x ) = √ x 2 + 2; x ∈ R f ◦ f ( x ) = x 4 + 6 x 2 + 12; x ∈ R g ◦ g ( x ) = p √ x-1-1; x ∈ [2 , ∞ ) 6. Sketch the graph and use the horizontal line test. See supplementary sheet. 1 7. f ( x ) = ( x-1) 2 , let x 1 = 2 ,x 2 = 0, then we have f ( x 1 ) = f ( x 2 ) = 1 . Restrict x to [1 , ∞ ) Then f-1 ( x ) = √ x + 1 Notice f-1 ( x ) =-√ x + 1 does not fit into this interval. 8. (a) see supplementary sheet (b) Apply horizontal line test (c) ( π 2 , 3 π 2 ) or [0 , π 2 ) or (0 ,π ] or ( π, 2 π ]. 2...
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This note was uploaded on 01/02/2012 for the course MATH 0211 taught by Professor Chan during the Spring '10 term at HKU.

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Assignment2solution(0211) - ( x ) 5. f ◦ g ( x ) = x + 2;...

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