Business Calc Homework w answers_Part_23

Business Calc Homework w answers_Part_23 - Section 3.8 b...

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Unformatted text preview: Section 3.8 b f (x) x→ g(x) (b) lim x2 lim a lim x2 x→ lim x→ (c) lim x→ f (x) g(x) lim x→ x5 x dy at dx 1, which is 7. 2x 1 at the point (1, 2). The slope is x2 a2 x2 1 b a x2 7. The slope of L is the reciprocal of the slope of its reflection 1 x2 8. 1 7 a2 x 1 y gets reflected to become x since a2 b x a x→ lim y a2 x→ lim 6. The reflection of line L is the tangent line to the graph of a2 b x a x→ 111 Quick Review 3.8 2 a2 x2 1. Domain: [ 1, 1] 1 Range: s Section 3.8 Derivatives of Inverse Trigonometric Functions (pp. 157–163) Exploration 1 Finding a Derivative on an Inverse Graph Geometrically 1. The graph is shown at the right. It appears to be a one-toone function At 1: , 22 2 2. Domain: [ 1, 1] Range: [0, ] At 1: 0 3. Domain: all reals Range: At 1: , 22 4 4. Domain: ( [ 4.7, 4.7] by [ 3.1, 3.1] 4 2. f (x) 5x 2. The fact that this function is always positive enables us to conclude that f is everywhere increasing, and hence one-to-one. Range: 0, , 1] 2 2 At 1: 0 1 3. The graph of f is shown to the right, along with the graph of f. The graph of f 1 is obtained from the graph of f by reflecting it in the line y x. 5. Domain: all reals Range: all reals At 1: 1 6. f (x) y 3x y 8 3x x [ 4.7, 4.7] by [ 3.1, 3.1] 4. The line L is tangent to the graph of f 1 at the point (2, 1). y 8 8 3 Interchange x and y: y f 1 7. f (x) (x) y [ 4.7, 4.7] by [ 3.1, 3.1] y 5. The reflection of line L is tangent to the graph of f at the point (1, 2). 3 x x 8 3 x 8 3 3 x y x 5 5 3 5 Interchange x and y: y f [ 4.7, 4.7] by [ 3.1, 3.1] 1 x3 5 (x) 3 5 x [1, ) , x 1 . It is . y 7 112 Section 3.8 8. f (x) 8 x 8 y y x 5. dy ds 1 f 9. f (x) 2s 6. 2 x xy 3x 3)x 7. 2 x 3 1 4s 2s d csc 1 (x 2 dx 3 1 y (x 2 (x2 2x 1) x s 1 d (5s) 1 ds (5s)2 5s s2 1 1 dy dx x2 s 25s 2 1) d (x 2 1 dx 1)2 1) 2 4 2x 2 (x Note that the condition x x 2 x2 1) 2 0 is required in the last step. 2 3 x 8. x arctan 3 y x , 3 tan y 4s d sec 1 5s ds 2 2 (x) 10. f (x) 1 1) (2) 2 1 2 y3 y 1) dy ds Interchange x and y: 1 (2s 2 x f 1 d (2s ds 1 2 1 3x y (y 2s 8 x 8 x (x) 1) 1 Interchange x and y: y d sec 1 (2s ds 3 tan y, 2 y 2 1 d x csc 1 dx 2 dx 1 dx 2 x2 2 x 2 2 y 2 dy dx x2 x 4 2 Interchange x and y: y f 3 tan x, 1 (x) x 2 3 tan x, 2 9. 2 x dy dt 1 1 d sec 1 t dt 1 t 2 1 Section 3.8 Exercises 1. dy dx 1 d cos 1 (x 2) dx 1 x4 1 d2 (x ) dx (x ) x4 1 1 1 t2 1 dy dx 10. 1 1 d cos 1 x dx d 1 2 dx x 1 1 1 1 x 2 1 x2 0 is required in the last step. dy dt 1 d 3 sin 1 2 dt t 1 x 1 1 1 x2 x t2 1 11. 1 dy dt d cot 1 dt 3 t2 6 6 t3 9 t4 d 3 2 dt t2 1 2. 1 Note that the condition t 22 1 2x (2x) 12 t 1 t 1 t d1 dt t 2 t t4 1 t 1 ( 9 d t)2 dt t 1 3. 4. dy dt d sin 1 dt dy dt d sin 1 (1 dt 1 2t t2 1 2t ( 2 t)2 1 1 t) 1 (1 d ( dt 2 2 t) d (1 2 dt t) 1 t) 2 t (t 1) 2t 2 12. dy dt d cot 1 dt 1 1 t t 1 1 1 ( 1 12t 1 ) d dt 1) 2 (t 1 2t t 1 t 1 1 113 Section 3.8 13. dy ds d (s ds 1 (s) 2 d (cos 1s) ds s2) 1 ( 2s) 2 1 1 s (1 s) 1 14. dy ds d ds [sin 1 1 (2) 4x 2 1 (2x)]2 1 dy dx 4x 2 sec 2 x, the slope at The tangent line is given by y d sec 1 s ds 1 y 1 (2s) s2 (2x) 1 2 1 s2 1 1 sin 2 (2x)] s 19. (a) Since s2 2 s 2 1 dx 2 s2 2s 2d (2x)] 2 1 2 1 [sin [sin s 2 d [sin 1(2x)] 1 dx 1 s2 1 2 dy dx 1 s2)(1) 1 s s2 1 ( 18. 1 (b) Since s2 s 2x 1 2x 1 x2 1 , the slope at 1, s2 s dy 15. dx 1 d (tan 1 dx x 1 1 1 1 x2 2 2 1) x2 x x 2 1 1 x2 x y 1 f x Note that the condition x dy dx 1 d cot 1 x dx 5x 4 1 1) 4 1 . 2 , or 6x 2 1. Thus f (1) 3 and f (x) includes the point (1, 3) and (x) will include (3, 1) and the slope will be 1 . (We have 12 (x) is defined and differentiable at 5x 4 6x 2 1, which is never zero.) 21. (a) Note that f (x) sin x 3, which is always between 2 and 4. Thus f is differentiable at every point on the interval( , ) and f (x) is never zero on this interval, so f has a differentiable inverse by Theorem 3. d (tan 1 x) dx d1 dx x ) (3) 3. This is true by Theorem 3, because f (x) 1 is required in the last step. 1 1 and ( f 1 assumed that f 1 0 16. 1 1 + 12 . 1 . Thus, f 1(3) 12 x2 x is the slope of the graph is 12 at this point, the graph of 1 x x2 4 1 1 1 4 (b) Since the graph of y 1 1) (2x) x2 1 2 20. (a) Note that f (x) f (1) 12. d (csc 1 x) dx d ( 1)2 dx x2 ( 1 x 2 y 1 2. 4 1, or 4 1 (x 2 The tangent line is given by y ss , 1 is sec 2 1. 2 dy dx 4 1 1 x2 1 ( 1 1 x2 1 ) 1 x2 1 x2 (b) f (0) cos 0 3(0) f (0) sin 0 3 x2 1 1; 3 (c) Since the graph of y 1 f (x) includes the point (0, 1) 1 x2 1 and the slope of the graph is 3 at this point, the graph of y 1 1 0, x 1 x2 f 1 (x) will include (1, 0) and the slope will be 1 . Thus, f 1(1) 3 22. 0 The condition x 0 is required because the original function was undefined when x 0. 17. dy dx d (x sin 1 x) dx 1 (x) 1 sin 1 x x2 d ( dx (sin 1 1 x 2) [ 2 , 2 ] by [ 4, 4] 1 x)(1) 2 1 x2 ( 2x) (a) All reals (b) , 22 0 and (f 1 ) (1) 1 . 3 114 Section 3.8 22. continued (b) y (c) At the points x (d) k , where k is an odd integer. 2 y 2 (c) None, since 30. (a) y 0. x2 x 1 0 (b) y 1 d sec 1 x dx 0 2 (c) None, since 1 d csc 1 x dx 0. x2 x 1 2π 1 x is 1. (b) None, since sin –2π 31. (a) None, since sin undefined for x x 1 x is undefined for x (c) None, since –2 1 x is undefined for x (b) None, since cos 1 x is undefined for x 1 d sin x 1 sin2 x dx cos x which is (c) None, since 23. (a) v(t) (b) a(t) (c) 24. 1 which is always positive. 1 1 So cos t2 2t which is always negative. (1 t 2)2 x, x sin 1 sin (b) β 1 x x 2 . x α 2 d cos 1 (x) dx d dx sin 2 1 x 1 tan 1 So tan x, x cot cot (c) x2 1 d cot 1 x dx 1 d sin 1 (x) dx 1 0 25. 0. x cos 1 x2 1. α cos x is positive or negative. dx dt dv dt 1 1. β 1 1 depending on whether 1 d cos 1 x dx 33. (a) sin2 x 1 0. x2 1 32. (a) None, since cos d sin 1 (sin x) dx (e) f (x) 1 d sin 1 x dx 1. 1 1 x x 2 . β x α d tan 1 (x) dx 2 d 0 tan 1 (x) dx 1 1 x2 1 1 sec So sec 1 x x, csc csc 1 x 1 x 2 . 34. d 26. csc 1 (x) dx d dx 2 x (x) x 2 A B 1 C 2 (b) y 2 (c) None, since 28. (a) y 1 d sec 1 (x) dx 1 0 27. (a) y sec d tan 1 x dx 1 0. x2 1 0 (b) y (c) None, since 29. (a) y 2 d cot 1 x dx 1 1 x2 0. The “straight angle” with the arrows in it is the sum of the three angles A, B, and C. A is equal to tan 1 3 since the opposite side is 3 times as long as the adjacent side. B is equal to tan 1 2 since the side opposite it is 2 units and the adjacent side is one unit. C is equal to tan 1 1 since both the opposite and adjacent sides are one unit long. But the sum of these three angles is the “straight angle,” which has measure radians. Section 3.9 35. s 6. Fold 1 log4 x 15 15 log4 x 12 12 log4 x log4 x ln Fold 2 s 2 Fold 3 s s s tan 1, so tan 1 tan 1 ln 19 ln 3 5t ln 5 1 1 tan 2 tan 1 t ln 5 s Section 3.9 Derivatives of Exponential and Logarithmic Functions (pp. 163–171) Exploration 1 Leaving Milk on the Counter 55 and solve; 72 30(0.98)t 4. ln (x 2 ln (x dy dx d (2e x) dx 2e x dy dx d 2x (e ) dx e 2x dy dx d ex dx dy dx d e 5x dx e 5. dy dx d 2x/3 e dx e 2x/3 6. dy dx d e x/4 dx e dy dx d (xe 2) dx dx (e ) dx dy dx d 2x (x e ) dx d (xe x) dx (e x)(2x) tan x x 2e x 2 ln (x 2) ln x x 4 2 ln (x 2)(x 2) x2 2) 5. log2 (8x 5) log2 (23)x 5 log2 23x 15 2.71 Section 3.9 Exercises e x ln 7 4) ln 3 ln 2 ln 3 x d (2x) dx xd e dx 2e 2x ( x) 5x d dx x e ( 5x) d 2x dx 3 5e 5x 2 2x/3 e 3 , ln 8 ln 5 ) ln 3 (x 2)(e x) ln (0.98) Quick Review 3.9 3. ln (e ln 2) 7. 17 30 0.343 degrees/minute. tan x x ln 2 4. 30 ln (0.98) (0.98)t. At t e ln 7 1) ln 3 x(ln 3 28.114 ln (0.98) ln 2. 7x ln 2x 3. 17 30 The milk reaches a temperature of 55 F after about 28 minutes. x 1 1.50 2x ln 3x (x 1 8. ln 1. log5 8 ln (ln 5) ln 5 2. 17 ln 30 t ln (0.98) dy 5. dt dy dt ln 18 1. 55 17 30 (0.98)t t ln (ln 5) 3x 10. 2. The temperature of the room is 72 F, the limit to which y tends as t increases. 3. The milk is warming up the fastest at t 0. The second derivative y 30(ln(0.98))2(0.98)t is negative, so y (the rate at which the milk is warming) is maximized at the lowest value of t. ln 18 t 1. The temperature of the refrigerator is 42 F, the temperature of the milk at time t 0. 4. We set y 18 ln 5 18 ln ln 5 ln 5t 3. 2.68 18 5t 1 ln (12x 2) ln 19 From Exercise 34, we have tan ln 3x ln 19 x 9. ln x 3 19 x ln 3 2. 0 ln (4x 4) 1 and 2, so s 2 (x 3)(12x 2) 3x ln 3x If s is the length of a side of the square, then tan ln (12x 2) 3x 8. 5 ,x 4 ln 3x 7. 3 ln x s 15 12 3x 15 9. 10. x/4 d xe x 1 x/4 e 4 x 4 dx e2 ex [(x)(e x) (e x)(1)] ex dy dx d ex dx e dy dx d (x 2) e dx e (x ) xd dx 2 ( x) d2 (x ) dx e x 2x 2xe (x ) 2 115 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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