Calculus- Early Transcendentals, 2021a

570 Selected Exercise Answers
1.3.7 ( √ 2 + √ 6)/4
1.3.8 −(1 + √ 3)/(1 − √ 3)=2 + √ 3
1.3.11 t = π/2
√
2
1.4.1 1.
2
2. 3( √ x + h + 1+ √ x + 1)
1.4.2 1. −13/5
2. −1/2,3
3. (1 ± √ 13)/2
4. No real solutions
√
5. 3
1.4.3 Counter-examples may vary
1. x = 3
2. x = h = 1
3. x = y = 1
1.4.4 x + 3y − 13 = 0, or equivalents such as y = − 1 3 x + 13 3
1.4.5 (−∞,0] ∪ ( 1 3 ,3]
1.4.6 It is impossible for both x − 2and1− x to be
non-negative for the same real number x.
1.4.7 6x + 3h
1.4.8 −1/[(2x + 2h − 1)(2x − 1)]
1.4.9 −3
1.4.17 x =(y − 2)/(3 + 4y)
1.4.18 Q(x)=x + 1, R = −7
2.1.1 1. {x | x ∈ R}, i.e., all x
2. {x | x ≥ 3/2}
3. {x | x ≠ −1}
4. {x | x ≠ 1andx ≠ −1}
5. {x | x < 0}
6. {x | x ∈ R}, i.e., all x
7. {x | h − r ≤ x ≤ h + r}
8. {x | x ≥ 0}
9. {x |−1 ≤ x ≤ 1}
10. {x | x ≥ 1}
11. {x |−1/3 < x < 1/3}
12. {x | x ≥ 0andx ≠ 1}
13. {x | x ≥ 0andx ≠ 1}
2.1.2 A = x(500 − 2x), {x | 0 ≤ x ≤ 250}
2.1.3 V = r(50 − πr 2 ), {r | 0 < r ≤ √ 50/π}
2.1.4 A = 2πr 2 + 2000/r, {r | 0 < r < ∞}
2.2.3 {x | x ≥ 3}, {x | x ≥ 0}
2.3.1 y = 2 x
2.3.2 y = 7
2.3.3 y = 2
2.3.4 x ≠ 0
1.4.10 π/6, 5π/6
1.4.11 2π/5
1.4.12 It is equal to 2 for all x larger than 4.
2.6.1 1. π/3
2. 3π/4
2.6.2 1. π/4
2. −π/3
3. 1/3
4. −3/4
1.4.13 (x + 2) 2 +(y − 3) 2 = 25.
1.4.14 Centre is (−3,2) and radius is 2
1.4.15 y could be any real number greater than or equal to 6
1.4.16 x 6 y 4 /(36z 8 )
2.6.3 √ 1 − x 2 /x with domain [−1,0) ∪ (0,1]
2.8.1 (d)
2.8.2 3/[5(x + 3)]
2.8.3 6 + h