Solution of Calculus_6e

# Solution of Calculus_6e - a 4 a 2 1 a 2(c f √ a = q 1 √...

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Section 1.1 C01S01.001: If f ( x )= 1 x , then: (a) f ( a )= 1 a = 1 a ; (b) f ( a 1 )= 1 a 1 = a ; (c) f ( a )= 1 a = 1 a 1 / 2 = a 1 / 2 ; (d) f ( a 2 )= 1 a 2 = a 2 . C01S01.002: If f ( x )= x 2 + 5, then: (a) f ( a )=( a ) 2 +5= a 2 +5; (b) f ( a 1 )=( a 1 ) 2 +5= a 2 +5= 1 a 2 +5= 1+5 a 2 a 2 ; (c) f ( a )=( a ) 2 +5= a +5; (d) f ( a 2 )=( a 2 ) 2 +5= a 4 +5. C01S01.003: If f ( x )= 1 x 2 +5 , then: (a) f ( a )= 1 ( a ) 2 +5 = 1 a 2 +5 ; (b) f ( a 1 )= 1 ( a 1 ) 2 +5 = 1 a 2 +5 = 1 · a 2 a 2 · a 2 +5 · a 2 = a 2 1+5 a 2 ; (c) f ( a )= 1 ( a ) 2 +5 = 1 a +5 ; (d) f ( a 2 )= 1 ( a 2 ) 2 +5 = 1 a 4 +5 . C01S01.004: If f ( x )= 1+ x 2 + x 4 , then: (a) f ( a )= p 1+( a ) 2 +( a ) 4 = 1+ a 2 + a 4 ; (b) f ( a 1 )= p 1+( a 1 ) 2 +( a 1 ) 4 = 1+ a 2 + a 4 = r ( a 4 ) · (1 + a 2 + a 4 ) a 4 = r a 4 + a 2 +1 a 4 = a 4 + a 2 +1 a 4 =
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Unformatted text preview: a 4 + a 2 + 1 a 2 ; (c) f ( √ a ) = q 1 + ( √ a ) 2 + ( √ a ) 4 = √ 1 + a + a 2 ; (d) f ( a 2 ) = p 1 + ( a 2 ) 2 + ( a 4 ) 2 = √ 1 + a 4 + a 8 . C01S01.005: If g ( x ) = 3 x + 4 and g ( a ) = 5, then 3 a + 4 = 5, so 3 a = 1; therefore a = 1 3 . C01S01.006: If g ( x ) = 1 2 x − 1 and g ( a ) = 5, then: 1...
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## This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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