Concepts Test

5 Concepts Test

1. False:

If fx () = x 3 , '( ) f x = 3 x 2 and the tangent line y = 0 at x = 0 crosses the

curve at the point of tangency.

2. False:

The tangent line can touch the curve

at infinitely many points.

= 4 x y 3 tan , which is unique for each

3. True:

value of x.

4. False:

m tan = – sin , x which is periodic.

5. True:

If the velocity is negative and

increasing, the speed is decreasing.

6. True:

If the velocity is negative and decreasing, the speed is increasing.

α = 0.5 7. True:

If the tangent line is horizontal, the

slope must be 0.

b ≠ . Then c fx ′ () = 2 ax = gx ′ ( ), but

f(x) ≠ g(x).

Dfgx x ( ( )) = fgxgx ′ ( ( )) ( ); x ′ since g(x) = x, gx ′ ( ) 1, = so Dfgx x ( ( )) = fgx ′ ( ( )).

Dy x = 0 because π is a constant, not

a variable.

11. True:

Theorem 3.2.A

12. True:

The derivative does not exist when the tangent line is vertical.

13. False:

( fg ⋅ )() ′ x = fxgx ()() ′ + gxfx ()() ′

14. True:

Negative acceleration indicates

decreasing velocity.

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15. True:

If fx () = xgx 3 ( ), then

29. True:

2 Dfx x () = xgx 3 ′ ()3 + xgx 2 () D x (sin ) x = – sin ; x

x [ xg x ′ ( ) 3 ( )]. + gx D x (sin ) x = – cos ; x

D x 4 (sin ) x = sin ; x

16. False:

Dy x = 3 x 2 ; At (1, 1):

D 5 x (sin ) x = cos x

m tan = 3(1) 2 = 3

Tangent line: y – 1 = 3(x – 1)

30. False:

D x (cos ) x = – sin ; x

D 2 (cos ) – cos ;

17. False: Dy

x = fxgx ()() ′ + gxfx ()() ′

Dy x = fxgx () ′′ () + ′

gxfx ′

x (cos ) ()() sin ;

gxf () ′′ () x + fxgx ′ ()() ′

D x (cos ) x = DD x [ x (cos )] x = D x (sin ) x

= fxgx () ′′ ()2()() + fxgx ′ ′ + f ′′ ()() xgx

Since D 13 + (cos )

x = D x (sin ), x

D n + 3 (cos ) x = D x n x (sin ). x

18. True:

The degree of y = ( x 3 + x ) 8 is 24, so

D 25 y = 0. tan x 1 sin x x 31. True: lim = lim

3 3 fx () gxfx ()()–()() ′ fxgx ′

= 15 t g 2 + which is greater 6

dt

21. True: hx ′ () = fxgx ()() ′ + gxfx

than 0 for all t.

hc ′ () = fcgc ()() ′ + gcfc ()() ′ 4

33. True:

V =π r 3

= f(c)(0) + g(c)(0) = 0

() =π 4 r

If 3, then

23. True: dr

hx ′ () = fgx ′ ( ( )) ⋅ gx ′ () 34. True:

When h > r, then

( f g ) (2) ′ = fg ′ ( (2)) ⋅ g ′ (2) 35. True:

V =π r , S =π 4 r

3 = f ′ (2) ⋅ g ′ (2) =⋅= 22 4 dV =π 4 r dr 2 =⋅ S dr

If Δr = dr, then dV =⋅Δ S r

26. False: Consider fx () = x . The curve always lies below the tangent.

36. False:

dy = 5 x dx 4 , so dy > 0 when dx > 0,

27. False: The rate of volume change depends

but dy < 0 when dx < 0.

on the radius of the sphere.

37. False:

The slope of the linear approximation

dr

is equal to

28. True:

= 4 fa '( ) = f '(0) =− sin(0)

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Sample Test Problems

⎡ 3 x 2 + 2 – 3( x + h )–21 2 ⎤

d. fx ′ () = lim ⎢ ⎜

2 – 2 ⎟ ⎟ ⎥ lim ⎢

2 2 ⎥ ⎣ ⎝ 3( x + h ) + 2 3 x + 2 ⎠ h ⎦ ⎥ h → 0 ⎢ ⎣ (3( x + h ) + 2)(3 x + 2) h ⎥ ⎦

h → 0 ⎣ ⎢ (3( x + h ) 2 + 2)(3 x 2 + 2) h ⎥ ⎦ h → 0 (3( x + h ) 2 + 2)(3 x 2 + 2)

sin[3( x + h )] – sin 3 x

sin(3 x + h 3 ) – sin 3 x

f. fx ′ () = lim

= lim

sin 3 (cos 3 –1) x = h = +

3sin 3 lim x

cos 3 lim x

(3sin 3 )(0) (cos 3 )3 lim x x

= (cos 3 )(3)(1) x = 3cos 3 x

cos[ ( π+ x h )] – cos π x

cos( π+π x h ) – cos π x

π sin π h lim – cos ⎞ ⎜ x ⎟ lim ⎜ sin x = (– cos π π x )(0) – ( sin π π=π x ) – sin π x

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( – )( t xt 2 ++ tx x 2 )(–) + t x

x 2 – t 2 Let v = t – x, then t = v + x and as

– 2sin 2 x

e. gx ′ () = lim = 2 cos 2 x ⋅ 0 – 2sin 2 1 –2sin 2 x ⋅= x t → x t – x

Other method:

( t – x )( t + x )

Use the subtraction formula

= lim

t → x 2sin( x ) sin( − ).

3. a. f(x) = 3x at x = 1

d. f(x) = sin x at x π =

Let v = t – x, then t = v + x and as t → xv , 0. →

f. f(x) = –sin 3x at x

sin π v cos π+ x sin π x cos π v – sin π x = lim

g. f(x) = tan x at x =

=π cos π⋅+π x 1 sin π⋅=π x 0 cos π x

h. fx () =

at x = 5

Other method:

Use the subtraction formula

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d ⎛ 4 x 2 2 3 9 –2 ⎞ ( x + x )(8 ) – (4 x x 2 – 2)(3 x 2 + 1)

e. [ f 2 ( )] t = 2()() ftft ′ =

At t = 2, f ′ ( (2)) (2) f f ′ = f ′ (2) (2) f ′ = –

6. D x ( x 3 –3 x 2 + x –2 ) = 3 x 2 –6 x + (–2) x –3 15. D (sin θ + cos θ 3 θ ) = cos θ + 3cos 2 θ (– sin ) θ

= 3 x 2 –6–2 x x –3 = cos – 3sin cos θ θ 2 θ

3 2 2 D 2 (sin θ + cos θ 3 θ ) 7. Dz z ( + 4 z + 2) z = 3 z + 8 z + 2

= – sin – 3[sin (2)(cos )(– sin ) cos θ θ θ θ + 3 θ ]

⎛ 3–5 x ⎞ ( x 2 + 1)(3) – (3 – 5)(2 ) x x

= – sin θ + 6sin 2 θ cos – 3cos θ 3 θ

8. D x ⎜ 2 ⎟ =

2 2 ⎝ x + 1 ⎠ ( x + 1)

d 2 2 2 − 3 x + 10 x + 3 16. [sin( ) – sin ( )] cos( )(2 ) – (2sin )(cos ) t t = t t t t =

dt

( x 2 + 1) 2

= t 2 cos( ) – sin(2 ) t 2 t

⎛ 4 t − 5 ⎞ (6 t 2 + t 2 )(4) – (4 – 5)(12 t t + 2)

9. D =

17. D [sin( θ 2 )] cos( = θ 2 )(2 ) θ = θ 2 cos( θ 2 t ⎜ 2 ⎟ 2 2 θ )

(6 t + 2) t

− 24 t 2 + 60 t + 10 d 3 = = 2 18. (cos 5 ) x (3cos 5 )(– sin 5 )(5) x x

(6 t 2 + 2) t 2 dx

= –15cos 5 sin 5 2 x x

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19. d [sin (sin( 2 π θ ))] = 2sin(sin( π θ )) cos(sin( π θ ))(cos( π θ ))( ) π =π 2 sin(sin( π θ )) cos(sin( π θ )) cos( π θ )

20. [sin (cos 4 )] 2 t = 2sin(cos 4 ) cos(cos 4 ) (– sin 4 )(4) t ( t ) t = –8sin(cos 4 ) cos(cos 4 ) sin 4 t t t

dt

21. D tan 3 (sec 3 )(3) 2 θ = 3sec 3 θ 2 θ = θ

d ⎛ sin 3 x ⎞ (cos 5 x 2 )(cos 3 )(3) – (sin 3 )(– sin 5 x x x 2 )(10 ) x 3cos 5 x 2 cos 3 x + x 10 sin 3 sin 5 x x 2

23. fx ′ () = ( x 2 –1) (9 2 x 2 – 4) (3 + x 3 – 4 )(2)( x x 2 –1)(2 ) x = ( x 2 –1) (9 2 x 2 – 4) 4 ( + xx 2 –1)(3 x 3 –4) x

f′ (2) = 672

24. gx ′ () = 3cos 3 x + 2(sin 3 )(cos 3 )(3) x x = 3cos 3 x + 3sin 6 x gx ′′ () = –9sin 3 x + 18cos 6 x g′′ (0) 18 =

d ⎛ cot x ⎞ (sec x 2 )(– csc 2 x ) – (cot )(sec x x 2 )(tan x 2 )(2 ) x – csc 2 x – 2 cot tan x x x 2

(cos – sin )(4 cos t t t t + 4sin ) – (4 sin )(– sin – cos ) t t t t t

26. D t ⎜

(cos – sin ) t t 2

t 4 cos 2 t + 2sin 2 – 4sin t 2 t + t 4 sin 2 t

4 t + 2sin 2 – 4sin t 2 t

= (cos – sin ) t t 2

(cos – sin ) t 2

27. fx ′ () = ( – 1) 2(sin x 3 π x – )( cos x π π x – 1) (sin + π x – ) 3( – 1) x 2 x 2 = 2( – 1) (sin x 3 π x – )( cos x π π x – 1) 3(sin + π x – ) ( – 1) x 2 x 2

f′ (2) 16 4 = −π≈ 3.43

28. ht ′ () = 5(sin(2 ) cos(3 )) (2 cos(2 ) – 3sin(3 )) t + t 4 t t ht ′′ () = 5(sin(2 ) cos(3 )) ( 4sin(2 ) – 9 cos(3 )) 20(sin(2 ) cos(3 )) (2 cos(2 ) – 3sin(3 )) t + t 4 − t t + t + t 3 t t 2 h′′ (0) =⋅⋅−+ 5 1 ( 9) 20 1 2 4 ⋅⋅ 3 2 = 35

29. gr ′ () = 3(cos 5 )(– sin 5 )(5) 2 r r = –15cos 5 sin 5 2 r r gr ′′ () = –15[(cos 5 )(cos 5 )(5) (sin 5 )2(cos 5 )(– sin 5 )(5)] 2 r r + r r r = –15[5cos 5 – 10(sin 5 )(cos 5 )] 3 r 2 r r

g ′′′ () r = –15[5(3)(cos 5 )(– sin 5 )(5) (10sin 5 )( sin 5 )(5) (cos 5 )(20sin 5 )(cos 5 )(5)] 2 r r − 2 r − r − r r r = –15[ 175(cos 5 )(sin 5 ) 50sin 5 ] − 2 r r + 3 r

g′′′ (1) ≈ 458.8

30. ft ′ () = hgtgt ′ ( ( )) ( ) 2 ( ) ( ) ′ + gtgt ′

31. Gx ′ () = Frx ′ (() + sx ( ))( ( ) rx ′ + sx ′ ( )) + sx ′ () Gx ′′ () = Frx ′ (() + sx ( ))( ( ) rx ′′ + sx ′′ ( )) ( ( ) + rx ′ + sxFrx ′ ( )) ′′ (() + sx ( ))( ( ) rx ′ + sx ′ ( )) + sx ′′ ()

= Frx ′ (() + sx ( ))( ( ) rx ′′ + sx ′′ ( )) ( ( ) + rx ′ + sx ′ ( )) 2 Frx ′′ (() + sx ( )) + sx ′′ ()

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32. Fx ′ () = QRxRx ′ ( ( )) ( ) ′ = 3[ ( )] (– sin ) Rx 2 x b. 128 – 16 t t 2 = 0

= –16t(t – 8) = 0

–3cos 2 x sin x

The object hits the ground when t = 8s v = 128 – 32(8) = –128 ft/s

33. Fz ′ () = rszsz ′ ( ( )) ( ) [3cos(3 ( ))](9 ′ = sz z 2 )

2 3 39. s = t 3 –6 t = 2 27 z cos(9 ) z + 9 t

3(t – 3)(t – 1) < 0

3(t – 3)(t – 1) = 0

35. V =π r 3

t = 1, 3

a (1) = –6, a(3) = 6

When r = 5,

=π 4 (5) 2 = 100 π≈ 314 m 3 per

40. a. D 20 ( x 19 + x 12 + x x 5 + 100) = 0 meter of increase in the radius.

c. D 20 x (7 x 21 + 3 x 20 ) =⋅ (7 21!) x +⋅ (3 20!)

d. D 20 x (sin x + cos ) x = D x 4 (sin x + cos ) x

When r = 5, dr 10 =π 4 (5) 2 = sin x + cos x dt

dr

0.0318 m/h

e. D 20 x 20 (sin 2 ) x = 2 sin 2 x

dt 10 π = 1,048,576 sin 2x

2 4 h 2 f. D 20 x ⎜⎟ ⎛⎞=

37. V = bh (12); ; = b = 1 (–1) (20!) 20 20!

When h = 3, = 9 18(3)

0.167 ft/min

v = 0, when t = 4 s dx 2 dy

= 2 = y + s 128(4) – 16(4)

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+ 3 xy 23 (1) 2 + 4(–2)(1) 8(1) + a. dy = – (–0.01)

y 2 –3 xy 32 ) = 3 xy 23 –3 x 2 = –0.0025

dx

2 dy 3 xy 23 –3 x 2 xy 23 – x 2 (–1) + 4(–2)(–1) 8(–1) = +

b. dy – 2(–2)(–1) 2(–2 2) 2 (–0.01) dx

d. x cos( xy ) ⎢ x + y ⎣ dx

⎥ + sin( ) xy = 2 x ⎦ d 45. a. [ f 2 () x + g 3 ( )] x

dx

x 2 cos( xy )

dy

= x 2 – sin( ) – xy xy cos( ) xy = 2()()3 fxfx ′ + g 2 ()() xgx ′

dx

f 2 (2) (2) 3 f ′ + g (2) (2) g ′

dy x 2 – sin( xy )– xy cos( ) xy

= dx

x 2 2 cos( xy ) = 2(3)(4) 3(2) (5) + = 84

2 ⎛ dy

e. x sec ( ) xy ⎜ x + y ⎟ + tan( xy ) = 0 b. [ ( ) ( )] fxgx = fxgx ()() ′ + gxfx ()() ′

⎝ dx

dx

f (2) (2) g ′ + g (2) (2) f ′ = (3)(5) (2)(4) + = 23

dy

x 2 sec ( ) 2 xy = –[tan( ) xy + xy sec ( )] 2 xy

dx

d c. [ ( ( ))]

fgx = fgxgx ′ ( ( )) ( ) ′

tan( ) xy + xy sec ( ) 2 xy

fg ′ ( (2)) (2) g ′ = f ′ (2) (2) g ′ = (4)(5) = 20

2 d. D x [ f 2 ( )] x = 2()() fxfx ′ 42. 2 yy

1 ′= 12 x

D 2 [ 2 2 x f ( )] x = 2[ ( ) fxf ′′ () x + fxfx ′ ( ) ( )] ′ 6x y 1 ′=

= f 2 (2) f ′′ (2) 2[ (2)] + f ′ 2

At (1, 2): y′ 1 = 3 = 2(3)(–1) 2(4) + 2 = 26

4 x + 6 yy′ 2 = 0

46. (13) = x + y ;2 =

Since ( )( y 1 ′ y 2 ′= ) –1 at (1, 2), the tangents are

When y= 5, x = 12, so

43. dy = [ cos( π π x )2] + x dx ; x = 2, dx = 0.01

= –4.8 ft/s

dy [ cos(2 ) 2(2)](0.01) = π π + = +

44. x (2 ) y + y 2 + y 2 [2( x + 2)] ( ++ x 2 2) (2) = 0 y = x sin15 °

= 400sin15 ° ≈ 104 mi/hr

When x = –2, y = ±1

⎛⎞− ⎜⎟ x x

b. D x x = D x ⎝⎠ ⎜ ⎟ =

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2 d. D 2

x ( x ) = D x (2 ) x = 2 x = 0, x = 1 or x = 2 The split points are 0, 1, and 2. The expression

sin θ

on the left can only change signs at the split

points. Check a point in the intervals ( −∞ ,0 ) , () 0,1 , () 1, 2 , and ( 2, ∞ ) . The solution set is

cos θ

b. D θ cos θ =

( sin ) − θ =− tan θ cos θ { xx | ≤ 0 or 1 ≤≤ x 2 } , or ( −∞ ,0 ][] ∪ 1, 2 .

50. a. fx () = x + 1; '( ) f x =− ( x + 1 ) ; a = 3

2 4. x 3 + 3 x 2 + 2 x ≥ 0

1 xx ( 3 x − 2 1/ 2 ) 0 = 4 +− (4)

Lx () = f (3) + f '(3)( x − 3)

( x − 3)

2 xx ( + 1 )( x + 2 ) ≥ 0

2 x +=− x +

xx ( + 1 )( x + 2 ) = 0

x = 0, x =− 1, x =− 2

b. fx () = x cos ; '( ) xfx =− x sin x + cos ; 1 xa = The split points are 0, − 1 , and − 2 . The expression on the left can only change signs at

Lx () = f (1) + f '(1)( x − 1)

the split points. Check a point in the intervals

= cos1 ( sin1 cos1)( +− + x − 1)

( −∞ − ,2 ) , ( −− 2, 1 ) , ( − ) 1, 0 , and ( 0, ∞ ) . The

= cos1 (sin1) − x + sin1 (cos1) + x − cos1

solution set is { x |2 −≤≤− x 1 or x ≥ 0 } , or

= (cos1 sin1) − x + sin1

≈− 0.3012 x + 0.8415

Review and Preview Problems

xx ( − 2 )

1. ( x − 2 )( x −< 3 ) 0 x 2 − 4

( x − 2 )( x − 3 ) = 0 xx ( − 2 )

2 or x = 3

( x − 2 )( x + 2 )

The split points are 2 and 3. The expression on The expression on the left is equal to 0 or the left can only change signs at the split points.

undefined at x = 0 , x = 2 , and x =− 2 . These

Check a point in the intervals ( −∞ ,2 ) , () 2,3 ,

are the split points. The expression on the left can only change signs at the split points. Check a

and ( 3, ∞ ) . The solution set is { x |2 << x 3 } or

point in the intervals: ( −∞ − ,2 ) , ( − 2, 0 ) , () 0, 2 , () 2,3 .

and ( 2, ∞ ) . The solution set is −2 −1 0 1 2 3 4 5 6 7 8 { xx | <− 2 or 0 ≤< x 2 or x > 2 } , or

2. x −−> x 6 0

( x − 3 )( x + 2 ) > 0 −5 −4 −3 −2 −1 0 1 2 3 4 5

( x − 3 )( x + 2 ) = 0

3 or x =− 2

The split points are 3 and − 2 . The expression on

the left can only change signs at the split points. Check a point in the intervals ( −∞ − ,2 ) , ( − 2,3 ) ,

and ( 3, ∞ ) . The solution set is

{ xx | <− 2 or x > 3 } , or ( −∞ − ∪ ,2 )( 3, ∞ ) .

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6. 2 > 0 is 0.

y ' = 2 tan x ⋅ sec 2 x ( x − 3 )( x + 3 )

2 > 0 2 tan sec x x = 0 x + 2

The expression on the left is equal to 0 at x = 3 ,

2sin x

and x =− 3 . These are the split points. The

cos x

expression on the left can only change signs at The tangent line is horizontal whenever the split points. Check a point in the intervals:

sin x = . That is, for x kπ 0 = where k is an

( −∞ − ,3 ) , ( − 3, 3 ) , and ( 3, ∞ ) . The solution set

integer.

is { xx | <− 3 or x > 3 } , or ( −∞ − ∪ ,3 )( 3, ∞ ) .

16. The tangent line is horizontal when the derivative is 0.

−5 −4 −3 −2 −1 0 1 2 3 4 5 y ' 1 cos =+ x The tangent line is horizontal whenever

3 3 7. f ' () x = 42 ( x + 1 )() 2 = 82 ( x + 1 ) cos x =− 1 . That is, for x = ( 2 k + 1 ) π where k is

an integer.

8. f ' () x = cos () π x ⋅= ππ cos () π x

17. The line y =+ 2 x has slope 1, so any line parallel

9. f ' () x = 2 ( x −⋅− 1 ) sin 2 () x ⋅+ 2 cos 2 ()() x ⋅ 2 x to this line will also have a slope of 1. For the tangent line to y =+ x sin x to be parallel

=− 2 ( x 2 − ) 1 sin 2 () x + x 2 cos 2 () x

to the given line, we need its derivative to equal 1.

y ' 1 cos =+ x = 1 x ⋅ sec tan x x − sec x ⋅ 1 cos x = 0

10. f ' () x =

The tangent line will be parallel to y =+ 2 x

sec xx ( tan x − 1 )

whenever x = ( 2 k + 1 ) .

2 18. Length:

11. f ' () x = ( 2 tan 3 x ) ⋅ sec 3 3 x ⋅ Width: −

− 24 2x

9 2x

= ( 6 sec 3 2 x ) ( tan 3 x )

Height: x