57314-0136147054_01

# 57314-0136147054_01 - CHAPTER 1 Section 1.1 1.1.1 If f(x = 1 then x 1 1(a f(a = = a a 1(b f(a1 = 1 = a a 1 1(c f a = = 1/2 = a1/2 a a(d f(a2 = 1 =

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CHAPTER 1 Section 1.1 1.1.1: 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 . 1.1.2: If f ( x x 2 + 5, then: (a) f ( a )=( a ) 2 +5= a 2 +5; (b) f ( a 1 a 1 ) 2 a 2 1 a 2 1+5 a 2 a 2 ; (c) f ( a a ) 2 a (d) f ( a 2 a 2 ) 2 a 4 +5. 1.1.3: If f ( x 1 x 2 +5 , then: (a) f ( a 1 ( a ) 2 = 1 a 2 ; (b) f ( a 1 1 ( a 1 ) 2 = 1 a 2 = 1 · a 2 a 2 · a 2 · a 2 = a 2 a 2 ; (c) f ( a 1 ( a ) 2 = 1 a ; (d) f ( a 2 1 ( a 2 ) 2 = 1 a 4 . 1.1.4: If f ( x 1+ x 2 + x 4 , then: (a) f ( a p 1+( a ) 2 +( a ) 4 = a 2 + a 4 ; (b) f ( a 1 p a 1 ) 2 a 1 ) 4 = 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 a 4 = a 4 + a 2 a 2 ; (c) f ( a q a ) 2 a ) 4 = a + a 2 ; (d) f ( a 2 p a 2 ) 2 a 4 ) 2 = a 4 + a 8 . 1.1.5: If g ( x )=3 x +4and g ( a )=5,then3 a +4=5,so3 a = 1; therefore a = 1 3 . 1

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1.1.6: If g ( x )= 1 2 x 1 and g ( a )=5,then: 1 2 a 1 =5; 1=5 · (2 a 1); 1=10 a 5; 10 a =6; a = 3 5 . 1.1.7: If g ( x x 2 +16and g ( a p a 2 +16=5; a 2 + 16 = 25; a 2 =9; a =3or a = 3 . 1.1.8: If g ( x x 3 3and g ( a )=5 ,then a 3 3=5,so a 3 = 8. Hence a =2. 1.1.9: If g ( x 3 x +25=( x + 25) 1 / 3 and g ( a )=5,then ( a + 25) 1 / 3 a +25=5 3 = 125; a = 100 . 1.1.10: If g ( x )=2 x 2 x +4and g ( a 2 a 2 a +4=5; 2 a 2 a 1=0; (2 a +1)( a 1) = 0; 2 a +1=0or a a = 1 2 or a =1 . 2
1.1.11: If f ( x )=3 x 2, then f ( a + h ) f ( a )=±3( a + h ) 2] [3 a 2] =3 a +3 h 2 3 a +2=3 h. 1.1.12: If f ( x )=1 2 x ,then f ( a + h ) f ( a )=±1 2( a + h )] [1 2 a ]=1 2 a 2 h 1+2 a = 2 h. 1.1.13: If f ( x )= x 2 f ( a + h ) f ( a )=( a + h ) 2 a 2 = a 2 +2 ah + h 2 a 2 =2 ah + h 2 = h · (2 a + h ) . 1.1.14: If f ( x x 2 x f ( a + h ) f ( a )=±( a + h ) 2 +2( a + h )] [ a 2 a ] = a 2 ah + h 2 a h a 2 2 a ah + h 2 h = h · (2 a + h +2) . 1.1.15: If f ( x 1 x f ( a + h ) f ( a 1 a + h 1 a = a a ( a + h ) a + h a ( a + h ) = a ( a + h ) a ( a + h ) = h a ( a + h ) . 1.1.16: If f ( x 2 x +1 f ( a + h ) f ( a 2 a + h 2 a = 2( a +1) ( a + h +1)( a 2( a + h ( a + h a = 2 a ( a + h a 2 a h ( a + h a = (2 a (2 a h ( a + h a = 2 a 2 a 2 h 2 ( a + h a = 2 h ( a + h a . 1.1.17: If x> 0then f ( x x | x | = x x =1 . 3

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If x< 0then f ( x )= x | x | = x x = 1 . We are given f (0) = 0, so the range of f is {− 1 , 0 , 1 } . That is, the range of f is the set consisting of the three real numbers 1, 0, and 1. 1.1.18: Given f ( x )=[ [3 x ]] , we s e e t h a t f ( x )=0 if 0 5 1 3 , f ( x )=1 1 3 5 2 3 , f (2) = 2 if 2 3 5 1; moreover, f ( x 3i f 1 5 2 3 , f ( x 2i f 2 3 5 1 3 , f ( x 1i f 1 3 5 0 . In general, if m is any integer, then f ( x )=3 m if m 5 x<m + 1 3 , f ( x m +1 m + 1 3 5 + 2 3 , f ( x m +2 m + 2 3 5 . Because every integer is equal to 3 m or to 3 m +1orto3 m + 2 for some integer m , we see that the range of f includes the set Z of all integers. Because f can assume no values other than integers, we can conclude that the range of f is exactly Z . 1.1.19: Given f ( x )=( 1) [ x ] , we Frst note that the values of the exponent [[ x ]] consist of all the integers and no other numbers. So all that matters about the exponent is whether it is an even integer or an odd integer, for if even then f ( x ) = 1 and if odd then f ( x 1. No other values of f ( x ) are possible, so the range of f is the set consisting of the two numbers 1and1 .
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## This document was uploaded on 11/28/2010.

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57314-0136147054_01 - CHAPTER 1 Section 1.1 1.1.1 If f(x = 1 then x 1 1(a f(a = = a a 1(b f(a1 = 1 = a a 1 1(c f a = = 1/2 = a1/2 a a(d f(a2 = 1 =

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