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Unformatted text preview: s constant 1 (b) y = xx , ln y = x ln x, y = 1 + ln x, y = xx (1 + ln x) y 57. f (x) = exe−1 (a) = 1 db/W/m2 10I0 ln 10 115 Chapter 4 58. Let P (x0 , y0 ) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and an equation of the tangent line at P is y − y0 = 3e3x0 (x − x0 ), y − e3x0 = 3e3x0 (x − x0 ). If the line passes through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0 e3x0 which gives x0 = 1/3 and thus y0 = e. The point is (1/3, e). 59. (a) (b) 60. (a) (b) f (x) = ln x; f (1) = lim h→0 ln(1 + h) − ln 1 ln(1 + h) 1 = lim = h→0 h h x d 10h − 1 = (10x ) h→0 h dx = 10x ln 10 f (x) = 10x ; f (0) = lim x=0 ln(e2 + h) − 2 d = (ln x) h→0 h dx dx 2x − 2 = (2 ) x→1 x − 1 dx 61. (b) = x=e2 = 2x ln 2 x=1 1 x = e−2 x=e2 = 2 ln 2 x=1 (c) y = ln 10 x=0 f (x) = ln x; f (e2 ) = lim f (x) = 2x ; f (1) = lim =1 x=1 6 dy 11 dy = − so &lt; 0 at dx 2x dx dy &gt; 0 at x = e x = 1 and dx 4 2 1 2 3 4 x (d) The slope is a continuous function which goes from a negative value to a positive value; therefore it must take the value zero in between, by the Intermediate Value Theorem. (e) 62. dy = 0 when x = 2 dx (a) (c) 100 -5 125 5 -5 5 -100 (d) 63. x = −2, 5/3 (a) ex cos πx oscillates between +ex and −ex as cos πx oscillates between −1 and +1. -25 y (b) 20 1 -20 2 3 x Exercise Set 4.5 64. (a) 116 12 0 9 0 (b) P tends to 12 as t gets large; lim P (t) = lim (c) 60 60 60 = = = 12 5 + 7e−t 5 5 + 7 lim e−t the rate of population growth tends to zero t→+∞ t→+∞ t→+∞ 3.2 0 9 0 65. (a) 100 0 8 20 (b) as t tends to +∞, the population tends to 19 95 95 95 = 19 = = lim P (t) = lim −t/4 t→+∞ t→+∞ 5 − 4e−t/4 5 5 − 4 lim e t→+∞ (c) the rate of population growth tends to zero 0 0 8 -80 EXERCISE SET 4.5 1. (a) −π/2 (b) π (c) −π/4 (d) 0 2. (a) π/3 (b) π/3 (c) π/4 (d) 4π/3 3. √ √ √ θ = −π/3; cos θ = 1/2, tan θ = − 3, cot θ = −1/ 3, sec θ = 2, csc θ = −2/ 3 117 4. Chapter 4 θ = π/3; sin θ = √ √ √ √ 3/2, tan θ = 3, cot θ = 1/ 3, sec θ = 2, csc θ = 2/ 3 5. tan θ = 4/3, 0 &lt; θ &lt; π/2; use the triangle shown to get sin θ = 4/5, cos θ = 3/5, cot θ = 3/4, sec θ = 5/3, csc θ = 5/4 5 4 θ 3 6. domain range sin−1 [−1, 1] [−π/2, π/2] −1 [−1, 1] [0, π ] tan−1 (−∞, +∞) (−π/2, π/2) −1 (−∞, +∞) (0, π ) cos cot sec−1 csc 7. −1 (−∞, −1] ∪ [1, +∞) [0, π/2) ∪ [π, 3π/2) (−∞, −1] ∪ [1, +∞) (0, π/2] ∪ (−π, −π/2] (b) sin−1 (sin π ) = sin−1 (sin 0) = 0 (a) π/7 (c) sin−1 (sin(5π/7)) = sin−1 (sin(2π/7)) = 2π/7 (d) Note that π/2 &lt; 630 − 200π &lt; π so sin(630) = sin(630 − 200π ) = sin(π − (630 − 200π )) = sin(201π − 630) where 0 &lt; 201π − 630 &lt; π/2; sin−1 (sin 630) = sin−1 (sin(201π − 630)) = 201π − 630. 8. (a) π/7 −1 (b) π −1 (c) (d) 9. 10. cos (cos(12π/7)) = cos (cos(2π/7)) = 2π/7 Note that −π/2 &lt; 200 − 64π &lt; 0 so cos(200) = cos(200 − 64π ) = cos(64π − 200) where 0 &lt; 64π − 200 &lt; π/2; cos−1 (cos 200) = cos−1 (cos(64π − 200)) = 64π − 200. (a) 0 ≤ x ≤ π (c) −π/2 &lt; x &lt; π/2 (b) −1 ≤ x ≤ 1 (d) −∞ &lt; x &lt; +∞ Let θ = sin−1 (−3/4) then sin θ = −3/4, −π/2 &lt; θ &lt; 0 and √ (see ﬁgure) sec θ = 4/ 7 ÷7 θ −3 4 11. Let θ = cos−1 (3/5), sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25 5 4 θ 3 Exercise Set 4.5 12. (a) 118 sin(cos−1 x) = 1 √ 1 − x2 (b) 1 − x2 tan(cos−1 x) = 1 cos−1x 1 − x2 cos−1x x √ (c) csc(tan−1 x) = 1 + x2 x 1+ x x2 (d) sin(tan−1 x) = √ 1 + x2 x tan−1x (a) cos(tan−1 x) = √ 1 1 + x2 (b) tan(cot−1 x) = 1 + x2 cot−1x x 1 √ x2 − 1 sin(sec−1 x) = x (d) cot(csc−1 x) = x2 − 1 x x 1 x2 − 1 csc−1x sec−1x x2 − 1 1 (a) 1 x 1 x tan−1x 14. x 1 1 + x2 (c) x 1 + x2 tan−1x 1 13. √ 1 − x2 x x −1.00 −0.80 −0.6 −0.40 −0.20 0.00 0.20 0.40 0.60 0.80 1.00 −1 sin x −1.57 −0.93 −0.64 −0.41 −0.20 0.00 0.20 0.41 0.64 0.93 1.57 cos−1 x 3.14 2.50 2.21 1.98 1.77 1.57 1.37 1.16 0.93 0.64 0.00 y (b) (c) y 3 2 1 1 x x 0.5 1 -1 1 119 15. Chapter 4 (a) y y π/2 3 x -5 1 5 -π/2 x -10 1 10 (b) The domain of cot−1 x is (−∞, +∞), the range is (0, π ); the domain of csc−1 x is (−∞, −1] ∪ [1, +∞), the range is [−π/2, 0) ∪ (0, π/2]. (a) y = cot−1 x, x = cot y , tan y = 1/x, y = tan−1 (1/x) (b) y = sec−1 x, x = sec y , cos y = 1/x, y = cos−1 (1/x) (c) y = csc−1 x, x = csc y , sin y = 1/x, y = sin−1 (1/x) 17. (a) 55.0◦ 18. (a) x = π − sin−1 (0.37) ≈ 2.7626 rad (b) θ = 180◦ + sin−1 (0.61) ≈ 217.6◦ 19. (a) x = π + cos−1 (0.85) ≈ 3.6964 rad (b) θ = − cos−1 (0.23) ≈ −76.7◦ 20. (a) x = tan−1 (3.16) − π ≈ −1.8773 (b) θ = 180◦ − tan−1 (0.45) ≈ 155.8◦ 21. (a) (b) −2/ 1 − (2x + 1)2 22. (a) 23. (a) 24. (a) y = 1/ tan x = cot x, dy/dx = − csc2 x (b) y = (tan−1 x)−1 , dy/dx = −(tan−1 x)−2 16. (b) 1 1...
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