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Solution Manual For Calculus, 6th Edition

Struggling with textbook exercises? Solution Manual For Calculus, 6th Edition breaks down solutions in a way that�s easy to understand.

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Section 1.1C01S01.001:Iff(x) = 1x, then:(a)f(−a) =1−a=−1a;(b)f(a−1) =1a−1=a;(c)f(√a) =1√a=1a1/2=a−1/2;(d)f(a2) =1a2=a−2.C01S01.002:Iff(x) =x2+ 5, then:(a)f(−a) = (−a)2+ 5 =a2+ 5;(b)f(a−1) = (a−1)2+ 5 =a−2+ 5 =1a2+ 5 = 1 + 5a2a2;(c)f(√a) = (√a)2+ 5 =a+ 5;(d)f(a2) = (a2)2+ 5 =a4+ 5.C01S01.003:Iff(x) =1x2+ 5 , then:(a)f(−a) =1(−a)2+ 5 =1a2+ 5 ;(b)f(a−1) =1(a−1)2+ 5 =1a−2+ 5 =1·a2a−2·a2+ 5·a2=a21 + 5a2;(c)f(√a) =1(√a)2+ 5 =1a+ 5 ;(d)f(a2) =1(a2)2+ 5 =1a4+ 5 .C01S01.004:Iff(x) =√1 +x2+x4, then:(a)f(−a) =√1 + (−a)2+ (−a)4=√1 +a2+a4;(b)f(a−1) =√1 + (a−1)2+ (a−1)4=√1 +a−2+a−4=√(a4)·(1 +a−2+a−4)a4=√a4+a2+ 1a4=√a4+a2+ 1√a4=√a4+a2+ 1a2;(c)f(√a) =√1 + (√a)2+ (√a)4=√1 +a+a2;(d)f(a2) =√1 + (a2)2+ (a4)2=√1 +a4+a8.C01S01.005:Ifg(x) = 3x+ 4 andg(a) = 5, then 3a+ 4 = 5, so 3a= 1; thereforea=13.C01S01.006:Ifg(x) =12x−1 andg(a) = 5, then:1

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Mathematics
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Calculus by C. Henry Edwards David ...

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