9. The x axis is a horizontal line so a 1 1 function can touch the x axis in at most one place.

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1 O Answers: Chapter 7 Contemporary Calculus PROBLEM ANSWERS Chapter Seven Section 7.0. f is one to one ( ), y is, g is not, h is not.. f is not, y is, g is, h is not. 5. I think SS numbers are supposeo be : each person shoul have one SS number, an each SS number shoul be assigneo only one person. Telephone numbers are not : the members of my family all have the same telephone number. 7. Monogamy. 9. The x axis is a horizontal line so a function can touch the x axis in at most one place.. (a) Yes. It passes the horizontal line test. (b) f(0) > f() so f is NOT increasing. f(0) < f(0.7) so f is NOT ecreasing.. f(x) ln(x). f '(x) x > 0 for x > 0. Assume a > b > 0 an f(a) f(b). Then by Rolle's theorem there is a point c between a an b so that f '(c) 0. But f '(c) 0 contraicts the fact that f '(x) > 0 for all x > 0, so our assumption that f(a) f(b) must be false. Therefore, f(a) f(b) an f is. 5. (a) Yes. (b) Yes. (c) ce () a b c e f (e) See Fig.. f e b a c 7. (a) Yes. (b) Yes. (c) fa () a b c e f (e) See Fig. 2. f e a c b (f) Same. (g) ENCODE( ENCODE( wor ) ) wor ENCODE( ENCODE( ba ) ) ENCODE( fa ) ba

2 O Answers: Chapter 7 Contemporary Calculus 2 Section 7.. See Fig... See Fig See Fig.. x f(x) f '(x) f (x) ( f ) '(x) Fig. : Complete Table 5 f '(2) 2 f '() f '() x h(x) h '(x) h (x) ( h ) '(x) h '() h '() h '(2) Fig. 2: Complete Table 7 unefine If (a, b) is on the graphs of both f an f, it is tempting, but wrong, to conclue that a b. For example, if f(x) x, then f (x) x anhe points (0,), (,0), an (, 2) all lie on both graphs, as o infinitely many other points. 9. f: (i) multiply by 7, (ii) subtract 5, an (iii) ivie by 4. f : (i) multiply by 4, (ii) a 5, an (iii) ivie by 7. f (x) 4x y g(x) 2x +. Interchange x an y: x 2y + so y x 2 an g (x) x 2. Checking: g ( g() ) g ( 2() + ) g ( ) () 2. (OK) g( g () ) g( (7) 2 ) g( ) 2() + 7. (OK). y w(x) 5 + ln(x). Interchange: x 5 + ln(y) so x 5 ln(y) an e x 5 e ln(y) y. w (x) e x 5. Check: w ( w() ) w ( 5 + ln() ) w ( 5 ) e 5 (5) e 0. (OK) 5. f goes through the point (,). ( f ) '() f '( f (x) ) f '( ) > f '(x) > 0 for all x means f is an increasing function. ( f ) '(x) f '( f (x) ) f '( number ) > 0 so f is also an increasing function.

3 O Answers: Chapter 7 Contemporary Calculus 9. Suppose f(x) x + b (b 0). Interchange: x y + b so y x b an f (x) x b. f an f both have slope so their graphs are parallel lines. The functions iffer by a constant, 2b, so the graphs o not intersect. 2. f(x) x + sin(x). f '(x) + cos(x) > 0 for all x so f is an increasing function an has an inverse. 2. (a) f (x) x f(x). (b) f (x) f(x) (c) f (x) f(x) () f (x) f(x) unless a 0 (e) f (x) f(x) 25. See Fig. 4. (a) unfol along the roa line to get B* (b) unfol along the river line to get A* Then the straight line from A* to B* is the shortest path from A* to B*. Get the final path by refoling the A* to B* line along the river an along the roa. This metho also works when the river an roa are not perpenicular to each other. Try a couple of your own esign. 27. Fol along the wall: the ball will lanhe same istance from the wall on your sie as the istance it woul have lane from the wall on the far sie. See Fig. 5.

4 O Answers: Chapter 7 Contemporary Calculus 4 Section 7.2. (a) sin(θ) so θ π 2 + 2πk. When k 0, θ π 2. When k, θ 5π 2. When k 2, θ 9π 2. (b) arcsin() π 2. arcsin() arcsin( ) π 2.. (a) sin(x) 0. when x 0.04, x π , an x 2π (b) sin(x) 0.4 when x.55 an x (c) sin(x) 0.5 when x 0.54, x π , an x 2π (a) tan(x).2 when x.268 an x π (b) tan(x) 0.2 when x an x π (a) sin(θ) 4 5 so θ arcsin( 4 5 ). (b) tan(θ) 4 so θ arctan( 4 ). (c) sec(θ) cos(θ) /5 5 so θ arcsec( 5 ). () cos(θ) 5 so θ arccos( 5 ). 9. (a) sin(α) 5 (b) tan(α) 5 2 (c) sec(α) cos(α) 2 () cos(α) 2.. See Fig.. θ arcsin( 2 7 ). (a) tan(θ) (b) cos(θ) 45 7 (c) csc(θ) sin(θ) 7 2 () cot(θ) See Fig. 2. θ arccos( 5 ). (a) tan(θ) 24 (b) sin(θ) 24 5 (c) csc(θ) sin(θ) 5 24 () cot(θ) See Fig.. θ arctan( a b ). (a) tan(θ) a b (b) sin(θ) a a 2 + b 2 (c) cos(θ) b a 2 + b 2 () cot(θ) b a. 7. sin 2 (θ) + cos 2 (θ) so sin(θ) cos 2 (θ) an cos(θ) sin 2 (θ). (a) sin( arccos(x) ) x 2 (b) cos( arcsin(x) ) x 2 (c) sec( arccos(x) ) x. 9. (a) No, arcsin(2) is not efine. arcsin() + arcsin() π 2 + π 2 π. (b) No, arccos(2) is not efine. arccos() + arccos()

5 O Answers: Chapter 7 Contemporary Calculus 5 2. See Fig. 4. Viewing angle θ α + β. (a) α arctan( 5 ) (.8o ) β arctan( 5 ) 0.97 (.o ) so θ ( 5.2 o ). (b) In general, if x is the istance in feet, then α arctan( x ) an β arctan( x ) so θ arctan( x ) + arctan( x ). 2. (a) See Fig. 5. y arrcsin( x 2 ). (b) See Fig. 6. y arctan( x 2 ). 25. sin(θ) h 20 so { sin(θ) } cos(θ). θ. Also, { sin(θ) } { h 20 } 20. h. Together, 20. h cos(θ). θ so h 20. cos(θ). θ. If θ. an θ 2, then h 20. cos(θ). θ 20. cos(. ). ( 2 ) cos(θ) h + 20 so { cos(θ) } sin(θ). θ. Also, { cos(θ) } { h + 20}. h. Together,. h sin(θ). θ so h. sin(θ). θ. If θ. an θ 2, then h. sin(θ). θ. sin(. ). ( 2 ).854.

6 O Answers: Chapter 7 Contemporary Calculus sin(θ) h 8 so { sin(θ) } cos(θ). θ. Also, { sin(θ) } { h 8 } 8. h. Together, 8. h cos(θ). θ so θ 8. cos(θ). h. If θ. an h 4, then θ 8. cos(θ). h 8.. ( 4 ) cos(. ). cos(θ) 7h 2 so { cos(θ) } sin(θ). θ. Also, { cos(θ) } { 7h 2 } 7. h. Together, 7. h sin(θ). θ so θ 7 sin(θ). h. If θ. an h 4, then θ 7 sin(θ). h 7 sin(. ). ( 4 ) tan(θ) h 4000 so { tan(θ) } sec2 (θ). θ. Also, { tan(θ) } { h 4000 } h. Together, h sec2 (θ). θ so h sec 2 (θ). θ. If θ π an θ π 2, then h sec 2 (θ). θ sec 2 ( π ). ( π 2 ) 4,88.8 ft/sec. 5. See Fig. 7. (a) sin(α) A C so α arcsin( A C ) (b) cos(β) A C so β arccos( A C ) (c) arcsin( A C ) + arccos( A C ) α + β π 2 ( or 90o ) 7. See Fig. 7. (a) sec(α) (b) csc(β) cos(α) C B so α arcsec( C B ). sin(β) C B so β arccsc( C B ). (c) arcsec( C B ) + arccsc( C B ) α + β π 2 ( or 90o ) 9. See Fig. 8. (a) tan(θ) so θ arctan( 5 2 ) (b) cot(θ) 2 5 so θ arccot( 2 5 ) 4. See Fig. 8. (a) sec(θ) cos(θ) so θ arcsec( 2 ) (b) csc(θ) sin(θ) so θ arccsc( 5 ). 4. arcsec() arccos( ).2 (or 70.5o )

7 O Answers: Chapter 7 Contemporary Calculus arcsec( ) arccos( ) π (or 80o ) 47. arccos(( 0.5 ) 2π ( or 20o ) 49. arccot( ) arctan( ) π ( or 45o ) 5. arccot( ) arctan( ) 0.22 ( or 8.4o ) 5. See Fig. 9. (a) sin(θ) 7 25 so θ arcsin( 7 25 ). (b) cos(θ) (c) arcsin( 7 25 so θ arccos( ) ) arccos( ) 55. If arcsec(x) α then sec(α) x so cos(α) x an cos(α) x. Since sin(α) cos 2 (α) x 2 x 2 x, then tan(α) sin(α) cos(α) ( x2 )/x /x x 2. tan(α) tan{ arcsec(x) } x 2. Finally, taking arctan of each sie of tan{ arcsec(x) } x 2, we get that arcsec( x ) arctan( x 2. (There are other ways to verify this result.) Section 7.. D( arcsin( x ) ). (x) 2 D( x ). D( arctan( x+5 ) ) + (x+5) 2. D( x+5 ) 5. D( arctan( x ) ) 7. D( ln( arctan(x) ) ) + ( x) 2. D( x ) 9x x 2 x 2 + 0x x. arctan(x). D( arctan(x) ) 4 x 2 2 x arctan(x). + x x x 4. 2 arcsin(x) x 2 9. D( ( arcsec(x) ) ) ( arcsec(x) ) 2. D( arcsec(x) ) ( arcsec(x) ) 2. x x x 2. D( arctan( ln(x) ) ) + ( ln(x) ) 2. D( ln(x) ). D( e x. arctan(2x) ) e x. D( arctan(2x) ) + arctan(x). D( e x ) e x. + ( ln(x) ) 2. x 2. x 2 4x + 4x 2. (2) + arctan(2x). e x.

8 O Answers: Chapter 7 Contemporary Calculus 8 4. arccos(x). x 2 arcsin(x). ( arccos(x) ) 2 x 2 5. D( arcsin(x) + arccos(x) ) x 2 + x x. + x 2 + arctan(x) 7. D( arcsin(x) ) D( ( arcsin(x) ) ). ( arcsin(x) ) 2. D( arcsin(x) ) ( arcsin(x) ) 2. (You coul also use the quotient rule.) x ( + arcsec(x) ) 2 x x D( sin( + arctan(x) ) ) cos( + arctan(x) ). D( + arctan(x) ) cos( + arctan(x) ). 20. tan(x). + x 2 + arctan(x). sec 2 (x). 2. D( x. arctan( x ) ) x. x. + ( /x ) 2. D( x ) + arctan( x ). D( x ) + ( /x ) 2 ( x 2 ) + arctan( x ) x x arctan( x ). + x arcsin( x ) + C 2. 0 x x 25 0 (x/5) 2 + x Put u x 5. Then u 5 x an x 5 u. 25 (u) u 5 arctan( u ) 5 arctan( x 5 ) 0 5 arctan( 5 ) 5 arctan( 0 5 ) arcsec( x 4 ) arcsec( 7 4 ) 5 4. arcsec( 5 4 ) x 2 x (x/7) 2 x Put u x 7. Then u 7 (u) 2 7 u 9 arcsin( u ) + C 9 arcsin( x 7 ) + C. x an x 7 u arctan( x 7 ) arctan( 4 7 ) 2 7. arctan( 4 7 ) 0.472

9 O Answers: Chapter 7 Contemporary Calculus x 5 ( x 5 ) x u (u) 2 Put u x 5. Then u 5 x an x 5 u. 5 u 5 arcsec( u ) 5 arcsec( x 5 ) 0 arcsec( 0 5 ) 5 arcsec( 6 5 ) 5 arccos( 5 0 ) 5 arccos( 5 6 ) arctan( x 5 ) + C. (Put u x 5 ) 29. e x + e 2x x. Put u ex. Then u e x x. + (u) 2 u arctan( u ) arctan( e x ) arctan( e ) arctan( e ) arctan( ln(e) ). arctan( ln() ) (Put u ln(x) ). cos(x) 9 sin 2 (x) x Put u sin(x). Then u cos(x) x. u 2 u arcsin( u ) + C arcsin( sin(x) ) + C. 2. 8x 6 + x 2 x Put u 6 + x 2. Then u 2x x an 4 u 8x x. u 4 u 4 ln( u ) + C 4. ln( 6 + x 2 ) + C.. 6x 9 + x 4 x Put u x 2. Then u 2x x an u 6x x. 9 + u 2 u.. arctan( u ) + C arctan( x2 ) + C. 5. (a) See Section 7.2, Problem 9: θ arctan( x ) + arctan( x ). (b) θ x + (/x) 2. x (/x) 2. x 2 x x which is never equal to 0, so the angle θ is maximize at an enpoint: θ is maximum when x 0. When x 0, θ π.

10 O Answers: Chapter 7 Contemporary Calculus 0 7. Separable. y y x x 2 so y y x. Then ln( y ) arcsin( x ) + C. 2 x Using the inital conition y(0) e: ln( e ) arcsin( 0 ) + C so 0 + C an C. ln( y ) arcsin( x ) + so ( e raiseo each sie) e ln( y ) e arcsin( x ) + an y e arcsin( x ) e. 8. Similar to 7: y e {arcsin( x/4 ) π/2} EXP( arcsin( x 4 ) π 2 ). 9. Separable: y y 2 x 9 + x 2 so y 2 y 9 + x 2 x an y arctan( x ) + C. Using the inital conition y() 2: 2 arctan( ) + C so C 2 arctan( ). Then y arctan( x ) + 2 arctan( ) an y arctan( x ) + 2 arctan(. ) 4. cos( arccos(x) ) x so D( cos( arccos(x) ) ) D( x ) an sin( arccos(x) ). D( arccos(x) ). Then D( arccos(x) ) sin( arccos(x) ) x 2. (Use the triangle with HYP, ADJ x, an OPP x 2 to verify that sin( arccos(x) ) x 2.) x 4. A(x) 0 + t2 t arctan( x ). (a) A(0) arctan(0) 0 (b) A() arctan() π/ (c) A(0) arctan(0).47. (b) (c) () lim x "# A(x) lim x "# arctan(x) π 2. x A(x) x arctan(x) + x 2 > 0. (We can also get A '(x) using the Funamental Thm. of Calculus.) Since A '(x) > 0 for all x, A(x) is an INCREASING function. 44. (a) 2. arctan(0) (b) 2( π 2 ) π x 5 x x 8x x x 5 x x 4. ln( x ) 5. arctan( x ) + C. 46. arctan( x ) 2. ln( x 2 + ) + C x + x x 7x x x + x x 7 2. ln( x ) + 0. arctan( x 0 ) + C ln( x ) + 5. arctan( x ) + C.

11 O Answers: Chapter 7 Contemporary Calculus 49. (a) Put u x +, u x. 8 (x+) 2 + x 8 u 2 + u 8. arctan( u ) + C 8. arctan( x+ ) + C. (b) Put u x 2 + 6x + 0. Then u (2x + 6) x an 2 u (4x + 2) x. 4x + 2 x 2 + 6x + 0 x 2 u u 2. ln u + C 2. ln( x 2 + 6x + 0 ) + C. (c) final result (answer in part a) + (answer in part b) 50. (a) 7. arctan( x+2 ) + C (b) 6. ln( x 2 + 4x + 5 ) + C (c) (answer in part a) + (answer in part b) 7. arctan( x+2 ) + 6. ln( x 2 + 4x + 5 ) + C 5. 6x + 5 x 2 + 4x + 20 x Put u x2 + 4x Then u (2x + 4) x an u (6x + 2) x. 6x + 2 x 2 + 4x + 20 x + x 2 + 4x + 20 x { a } + { b } { a } 6x + 2 x 2 + 4x + 20 x u u. ln u. ln( x 2 + 4x + 20 ) { b } (x+2) x (u) x 4 arctan( x+2 4 ) + C (Put u x+2 an u x) Total integral { a } + { b }. ln( x 2 + 4x + 20 ) + 4 arctan( x+2 4 ) + C. 52. ln( x 2 4x + ) +. arctan( x 2 ) + C.