Calculus with Analytic Geometry by edwards & Penney soln ch1

# Calculus with Analytic Geometry

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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 ) = 1 + ( a ) 2 + ( a ) 4 = 1 + a 2 + a 4 ; (b) f ( a 1 ) = 1 + ( a 1 ) 2 + ( a 1 ) 4 = 1 + a 2 + a 4 = ( a 4 ) · (1 + a 2 + a 4 ) a 4 = a 4 + a 2 + 1 a 4 = a 4 + a 2 + 1 a 4 = a 4 + a 2 + 1 a 2 ; (c) f ( a ) = 1 + ( a ) 2 + ( a ) 4 = 1 + a + a 2 ; (d) f ( a 2 ) = 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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1 2 a 1 = 5; 1 = 5 · (2 a 1); 1 = 10 a 5; 10 a = 6; a = 3 5 . C01S01.007: If g ( x ) = x 2 + 16 and g ( a ) = 5, then: a 2 + 16 = 5; a 2 + 16 = 25; a 2 = 9; a = 3 or a = 3 . C01S01.008: If g ( x ) = x 3 3 and g ( a ) = 5, then a 3 3 = 5, so a 3 = 8. Hence a = 2. C01S01.009: If g ( x ) = 3 x + 25 = ( x + 25) 1 / 3 and g ( a ) = 5, then ( a + 25) 1 / 3 = 5; a + 25 = 5 3 = 125; a = 100 . C01S01.010: If g ( x ) = 2 x 2 x + 4 and g ( a ) = 5, then: 2 a 2 a + 4 = 5; 2 a 2 a 1 = 0; (2 a + 1)( a 1) = 0; 2 a + 1 = 0 or a 1 = 0; a = 1 2 or a = 1 . C01S01.011: 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. C01S01.012: If f ( x ) = 1 2 x , then 2
f ( a + h ) f ( a ) = [1 2( a + h )] [1 2 a ] = 1 2 a 2 h 1 + 2 a = 2 h. C01S01.013: If f ( x ) = x 2 , then 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 ) . C01S01.014: If f ( x ) = x 2 + 2 x , then f ( a + h ) f ( a ) = [( a + h ) 2 + 2( a + h )] [ a 2 + 2 a ] = a 2 + 2 ah + h 2 + 2 a + 2 h a 2 2 a = 2 ah + h 2 + 2 h = h · (2 a + h + 2) . C01S01.015: If f ( x ) = 1 x , then 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 ) . C01S01.016: If f ( x ) = 2 x + 1 , then f ( a + h ) f ( a ) = 2 a + h + 1 2 a + 1 = 2( a + 1) ( a + h + 1)( a + 1) 2( a + h + 1) ( a + h + 1)( a + 1) = 2 a + 2 ( a + h + 1)( a + 1) 2 a + 2 h + 2 ( a + h + 1)( a + 1) = (2 a + 2) (2 a + 2 h + 2) ( a + h + 1)( a + 1) = 2 a + 2 2 a 2 h 2 ( a + h + 1)( a + 1) = 2 h ( a + h + 1)( a + 1) . C01S01.017: If x > 0 then f ( x ) = x | x | = x x = 1 . If x < 0 then 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. C01S01.018: Given f ( x ) = [[3 x ]], we see that 3

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f ( x ) = 0 if 0 x < 1 3 , f ( x ) = 1 if 1 3 x < 2 3 , f (2) = 2 if 2 3 x < 1; moreover, f ( x ) = 3 if 1 x < 2 3 , f ( x ) = 2 if 2 3 x < 1 3 , f ( x ) = 1 if 1 3 x < 0 . In general, if m is any integer, then f ( x ) = 3 m if m x < m + 1 3 , f ( x ) = 3 m + 1 if m + 1 3 x < m + 2 3 , f ( x ) = 3 m + 2 if m + 2 3 x < m + 1 . Because every integer is equal to 3 m or to 3 m + 1 or to 3 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 .
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