1

General Trigonometric Functions Answer Key 11-1 Free Response Homework

1. ddx

cos4 x( ) = ddx

cos x( )4⎡⎣ ⎤⎦

= 4 cos x( )3 ⋅−sin x= −4cos3 xsin x

3. ddx

sin2 x + cos2 x( ) = ddx1( )

= 0

5. ddx

cos4 2x( ) = ddx

cos2x( )4⎡⎣ ⎤⎦

= 4 cos2x( )3 ⋅−sin2x ⋅2= −8cos3 2xsin2x

7.

ddx

sec x2

3⎛⎝⎜

⎞⎠⎟= sec x

2

3tan x

2

3⋅ 23x

= 23xsec x

2

3tan x

2

3

9.

ddx

csc x43( ) = ddx

csc x43( )

= −csc x43 cot x

43 ⋅ 43x13

= − 43x13 csc x

43 cot x

43

= − 43

x3 csc x43 cot x43

2

11.

′f x( ) = sec2 x + sec x tan x

′fπ3

⎛⎝⎜

⎞⎠⎟ = sec

2 π3+ secπ

3tanπ

3

= 2( )2 + 2 ⋅ 3

= 4 + 2 3

13.

′f x( ) = 32

+ cos x = 0, DNE, EoaASD

c.v. : x = ± 5π6

,±π

15.

′f x( ) = 32

− sin x = 0, DNE, EoaASD

c.v. : x = π3

, 2π3

,0

17. v t( ) = 2 sin t( )cos t

= 2sin t cos t= sin2t

3

19.

′H t( ) = 4 ⋅−sin π8t − 5( )⎡

⎣⎢⎤⎦⎥⋅ π8

= − π2

sin π8t − 5( )⎡

⎣⎢⎤⎦⎥= 0, DNE, EoaASD

sin π8t − 5( )⎡

⎣⎢⎤⎦⎥= 0

π8t − 5( ) = 0 ± 2πn

π ± 2πn⎧⎨⎩

= 0 ±πn{

t − 5 = 0 ±16n8 ±16n

⎧⎨⎩

= 0 ± 8n{

t = 5 ±16n13±16n

⎧⎨⎩

= 5 ± 8n{

c.v. : t = 5, 13, 0, 20

vt 0 5 13 20← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

END + 0 − 0 + END

The weight moves up during the first 5 seconds, then back down until 13 seconds have passed, the back up again until time reaches 20 seconds. 11-1 Multiple Choice Homework

1. B limx→0

1− cos xx2

= 00

=LHlimx→0

sin x2x

= 00

=LHlimx→0

cos x2

= 12

4

3. E y = cos x( )2 − sin x( )2

′y = 2 cos x( ) ⋅sin x − 2 sin x( ) ⋅cos x= −4cos x ⋅sin x

5. D For f to be continuous at x = π : 1. f π( ) = ?

2. limx→π

f x( ) = limx→π

f x( ) = limx→π

tan xsin x

= 00

=LH

limx→π

sec2 xcos x

= 1−1

= −1

3. f π( ) = limx→π

f x( )∴ f π( ) = −1

11-2 Free Response Homework

1. ddx

x3 ⋅sec x( ) = x3 ⋅sec x tan x + sec x ⋅3x2= x2 sec x x tan x + 3( )

3. Dx x2 ⋅sin x + 2x ⋅cos x( )= x2 ⋅cos x + sin x ⋅2x + 2x ⋅−sin x + cos x ⋅2= x2 cos x + 2xsin x − 2xsin x + 2cos x

= x2 cos x + 2xsin x −2xsin x + 2cos x= x2 cos x + 2cos x

= cos x x2 + 2( )

5

5. ′f x( ) = 2 sin2 x ⋅2cos x ⋅−sin x + cos2 x ⋅2sin x ⋅cos x⎡⎣ ⎤⎦

= 2 −2sin3 xcos x + 2cos3 xsin x⎡⎣ ⎤⎦= 4sin xcos x −sin2 x + cos2 x⎡⎣ ⎤⎦= 4sin xcos x cos2 x − sin2 x⎡⎣ ⎤⎦= 2 ⋅2sin xcos x cos2x[ ]= 2sin2x ⋅cos2x= sin 4x

7. dydx

=cos x − 3( )sec2 x − tan x ⋅−sin x

cos x − 3( )2

=cos x − 3( ) ⋅ 1

cos2 x+ sin

2 xcos x

cos x − 3( )2

=

cos x − 3cos2 x

+ sin2 x ⋅cos xcos2 x

cos x − 3( )2

= cos x − 3+ sin2 xcos x

cos2 x cos x − 3( )2

=cos x 1− 3sec x + sin2 x( )cos2 x cos x − 3( )2

=cos x 1− 3sec x + sin2 x( )cos 2 x cos x − 3( )2

= 1− 3sec x + sin2 x

cos x cos x − 3( )2

6

9. ddx

sin x1− cos x

⎛⎝⎜

⎞⎠⎟ =

1− cos x( )cos x − sin x ⋅sin x1− cos x( )2

= cos x − cos2 x − sin2 x

1− cos x( )2

=cos x − cos2 x + sin2 x( )

1− cos x( )2

= cos x −11− cos x( )2

=− 1− cos x( )1− cos x( )2

=− 1− cos x( )1− cos x( ) 2

= −11− cos x

= − 1

2 1− cos x2

⎛⎝⎜

⎞⎠⎟

= − 1

2sin2 12x

= − 12csc2 1

2x

11.

′f x( ) = sec x ⋅sec2 x + tan x ⋅sec x tan x

′fπ4

⎛⎝⎜

⎞⎠⎟ = sec

π4⋅sec2 π

4+ tanπ

4⋅secπ

4tanπ

4

= 2 ⋅ 2( )2 +1⋅ 2 ⋅1

= 2 2 + 2 = 3 2

7

13.

′f x( ) = x ⋅−sin x + cos x + x ⋅cos x + sin x

′fπ4

⎛⎝⎜

⎞⎠⎟ =

π4⋅−sinπ

4+ cosπ

4+ π4⋅cosπ

4+ sinπ

4

= π4⋅− 1

2+ 1

2+ π4⋅ 12+ 1

2

= π4⋅− 1

2+ 1

2+ π4⋅ 12

+ 12

= 22= 2

15.

Volume = l Areatrapezoid( )= l 1

2b1 + b2( )h⎡

⎣⎢⎤⎦⎥

= 12 12

3+ 3+ 2y( ) 4 − y2⎡⎣⎢

⎤⎦⎥

= 12 12

6 + 4sin x( ) 4 − 4sin2 x⎡⎣⎢

⎤⎦⎥

= 12 12

6 + 4sin x( ) ⋅2 ⋅ 1− sin2 x⎡⎣⎢

⎤⎦⎥

= 12 12

6 + 4sin x( ) ⋅ 2 ⋅ 1− sin2 x⎡

⎣⎢

⎤

⎦⎥

= 12 6 + 4sin x( ) cos2 x⎡⎣

⎤⎦

= 12 6 + 4sin x( )cos x⎡⎣ ⎤⎦= 72cos x + 48sin xcos x

Volum ′e = −72sin x + 48 sin x ⋅−sin x + cos x ⋅cos x[ ]= −72sin x − 48sin2 x + 48cos2 x

c.v. : x = 0.439 radians ≡ 25.175!

122

3

2x x

y y

Note:

sin x = y2

y2 + h2 = 4⇒ h = 4 − y2h h

8

17 a.

′x t( ) = 5cos t, ′y t( ) = 2t!v = 5cos t, 2t b.

′′x t( ) = −5sin t, ′′y t( ) = 2!a = −5sin t, 2 c. Speed t=π = 5cosπ( )2 + 2π( )2 = 8.030 19. ′x t( ) = et ⋅cos t + sin t ⋅et

′y t( ) = et ⋅−sin t + cos t ⋅et′x 1( ) = ecos1+ esin1′y 1( ) = −esin1+ ecos1

Speed t=1 = ecos1+ esin1( )2 + −esin1+ ecos1( )2

= e2 cos21+ e2 sin1cos1+ e2 sin21+ e2 sin21− e2 sin1cos1+ e2 cos21

= e2 cos21+ e2 sin1cos1 + e2 sin21+ e2 sin21−e2 sin1cos1 + e2 cos21

= e2 cos21+ e2 sin21+ e2 sin21+ e2 cos21

= e2 cos21+ sin21+ sin21+ cos21( )= e2 1+1( )= e 2

11-2 Multiple Choice Homework

1. B

Min @ x ≈ −1.253 , Max @ x ≈1.253 No sign change @ x = 0

′f x( )

9

3. B ′y = x ⋅−sin x + cos x = 0

⇒ xsin x = cos x⇒ x = cot x

⇒ tan x = 1x

5. E ′h x( ) = f x( ) ⋅ ′g x( ) + g x( ) ⋅ ′f x( ) = f x( ) ′g x( )

⇒ g x( ) ⋅ ′f x( ) = 0⇒ g x( ) = 0 or ′f x( ) = 0We are given that g x( ) > 0, so ′f x( ) = 0.⇒ f x( ) = constantWe are given f 0( ) = 1, ∴ f x( ) = 1.

11-3 Free Response Homework

1.

′y = 1

1− x 2( )2⋅ 2

= 21− 2x2

3.

′y = − 1

1+ 1x

⎛⎝⎜

⎞⎠⎟2 ⋅−x

−2 − 11+ x2

= − 1

1+ 1x2

⋅− 1x2

− 11+ x2

= 1x2 +1

− 11+ x2

= 0

10

5.

′y =x ⋅ 1

x x2 −1− sec−1 x

x2

=

xx x2 −1

− sec−1 x ⋅ x2 −1x2 −1

x2

=

x − sec−1 x x2 −1x2 −1x2

= x − x2 −1sec−1 xx2 x2 −1

7. ′y = −1

2e3x 2e3x( )2 −1⋅2e3x ⋅3

= −1

2e3x 2e3x( )2 −1⋅ 2e3x ⋅3

= −34e6x −1

11

9.

′y = 12x1− x2

⎛⎝⎜

⎞⎠⎟2

+1⋅1− x2( ) ⋅2 − 2x ⋅−2x

1− x2( )2

= 1

4x2

1− x2( )2+1− x2( )21− x2( )2

⋅ 2 − 2x2 + 4x2

1− x2( )2

= 14x2 + 1− x2( )2

1− x2( )2⋅ 2 + 2x

2

1− x2( )2

=1− x2( )2

4x2 + 1− x2( )2⋅2 1+ x2( )1− x2( )2

=1− x2( )2

4x2 +1− 2x2 + x4⋅2 1+ x2( )1− x2( )2

=2 1+ x2( )x4 + 2x2 +1

=2 x2 +1( )

x2 +1( ) x2 +1( )

=2 x2 +1( )

x2 +1( ) x2 +1( )= 2x2 +1

12

11.

Speed t=4 = tan−1 45

⎛⎝⎜

⎞⎠⎟

2

+ ln17( )2 = 2.912

′′x t( ) = 1t

t +1⎛⎝⎜

⎞⎠⎟

2

+1⋅t +1( ) ⋅1− t ⋅1

t +1( )2 , ′′y t( ) = 1t 2 +1

⋅2t

′′x t( ) = 1t 2

t +1( )2 +1⋅ 1t +1( )2 , ′′y t( ) = 2t

t 2 +1

′′x t( ) = 1t 2

t +1( )2 +t +1( )2

t +1( )2

⋅ 1t +1( )2 , ′′y t( ) = 2t

t 2 +1

′′x t( ) = 1t 2 + t +1( )2

t +1( )2

⋅ 1t +1( )2 , ′′y t( ) = 2t

t 2 +1

′′x t( ) = t +1( )2

t 2 + t +1( )2 ⋅1

t +1( )2 , ′′y t( ) = 2tt 2 +1

′′x t( ) =t +1( )2

t 2 + t +1( )2 ⋅1

t +1( )2, ′′y t( ) = 2t

t 2 +1

′′x t( ) = 1t 2 + t +1( )2 , ′′y t( ) = 2t

t 2 +1

′′x 4( ) = 142 + 4 +1( )2 , ′′y t( ) = 2 ⋅4

42 +1

a t( ) = 841

, 817

11-3 Multiple Choice Homework

1. B Graph y = arcsin x and y = 2arccos x . Use the intersection menu item – x = 0.866 ≈ 0.9

13

3. E ′g x( ) = 1

1− 2x( )2⋅2

= 21− 4x2

5. D ddx

y2 + xy +1( )3 = 0⎡⎣ ⎤⎦

2y dydx

+ 3 xy +1( )2 ⋅ x dydx

+ y⎛⎝⎜

⎞⎠⎟ = 0

Plug in 2, −1( )

2 −1( ) dydx

+ 3 2 ⋅ −1( ) +1( )2⋅ 2 dy

dx+ −1( )⎛

⎝⎜⎞⎠⎟ = 0

−2 dydx

+ 3 −1( )2 ⋅ 2 dydx

−1⎛⎝⎜

⎞⎠⎟ = 0

−2 dydx

+ 3 2 dydx

−1⎛⎝⎜

⎞⎠⎟ = 0

−2 dydx

+ 6 dydx

− 3= 0

4 dydx

= 3

dydx

= 34

14

11- 4 Free Response Homework

1. ′y = sin 3x ⋅−sin 3x ⋅3+ cos 3x ⋅cos 3x ⋅3

= 3 cos2 3x ⋅−sin2 3x( )= 3cos6x = 0, DNE , EoaASD

3cos6x = 0cos6x = 0

6x =

π2± 2πn

− π2± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

= ± π2± 2πn⎧

⎨⎩

x =

π12

± π3n

− π12

± π3n

⎧

⎨⎪⎪

⎩⎪⎪

= ± π12

± π3n⎧

⎨⎩

3. ′y = 2cot 5x ⋅−csc2 5x ⋅5

= −10csc2 5xcot 5x = 0, DNE, EoaASD

csc2 5x = 0csc5x = 0sin5x = DNE⇒ no solution

or

cot 5x = 0tan5x = DNE

5x =

π2± 2πn

− π2± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

= π2±πn⎧

⎨⎩

x =

π10

± 2π5n

− π10

± 2π5n

⎧

⎨⎪⎪

⎩⎪⎪

= π10

± π5n⎧

⎨⎩

15

csc2 5x = DNEcsc5x = DNEsin5x = 0

5x = 0 ± 2πnπ ± 2πn

= 0 ±πn{⎧⎨⎩

x =0 ± 2π

5n

π5± 2π5n

= 0 ± 2π5n⎧

⎨⎩

⎧

⎨⎪⎪

⎩⎪⎪

or

cot5x = DNEtan5x = 0

5x = 0 ± 2πnπ ± 2πn

⎧⎨⎩

= 0 ±πn{

x =0 ± 2π

5n

π5± 2π5n

= 0 ± 2π5n⎧

⎨⎩

⎧

⎨⎪⎪

⎩⎪⎪

x = ± π10

± 2π5n, 0 ± 2π

5n

5.

′y = 32

− sin x = 0, DNE , EoaASD

sin x = 32

x =

π3± 2πn

2π3

± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

7. No fractions with a denominator of zero, no radicals, and no logarithms ∴ Domain: x∈ 0, 2π⎡⎣ ⎤⎦ Axis Points: −sin x = 0

x = 0 ± 2πnπ ± 2πn

⎧⎨⎩

= 0 ±πn{

⇒ x = 0, π , 2π

0, 0( ), π , π2

⎛⎝⎜

⎞⎠⎟ , 2π , π( )

h = 0

16

EB: None POEs: None VAs: None Extreme Points: ′y = 1

2− cos x = 0, DNE , EoaASD

12− cos x = 0

cos x = 12

x =

π3± 2πn

− π3± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

= π3±πn⎧

⎨⎩

c.v. : x = π3

, 2π3

, 0, 2π

π3 , −0.342⎛

⎝⎜⎞⎠⎟, 5π

3 , 3.484⎛⎝⎜

⎞⎠⎟,

0, 0( ), 2π , π( )

Range: y∈ −0.342, 3.484[ ]

−π/2 π/2 π 3π/2 2π

−4

−3

−2

−1

1

2

x

y

17

9. No fractions with a denominator of zero, no radicals, and no logarithms ∴ Domain: x ∈ 0, 2π[ ] Axis Points: sin x = 0

x = 0 ± 2πnπ ± 2πn

⎧⎨⎩

= 0 ±πn{

⇒ x = 0, π , 2π0, 0( ), π , 0( ), 2π , 0( )

h = 0 EB: None POEs: None

VAs: None Extreme Points: ′y = x ⋅cos x + sin x = 0, DNE , EoaASDxcos x + sin x = 0xcos x = −sin xxcos xcos x

= −sin xcos x

x = − tan xc.v. : x = 2.029, 4.914, 0, 2π

0, 0( ), 2π , 0( ) 4.913, − 4.814( ), 2.029, 1.820( )

Range: y∈ −4.814, 1.820[ ]

18

11. No fractions with a denominator of zero, no radicals, and no logarithms ∴ Domain: x ∈ −π , π[ ] h=0 EB: None POEs: None VAs: None Axis Points: sin x = 0

x = 0 ± 2πnπ ± 2πn

⎧⎨⎩

= 0 ±πn{

⇒ x = −π , 0, π−π , 0( ), 0, 0( ), π , 0( )

Extreme Points:

−π/2 π/2 π 3π/2 2π

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

x

y

19

′y = ex ⋅cos x + sin x ⋅ex ⋅= 0, DNE , EoaASDex cos x + sin x( ) = 0

ex = 0

ln ex( ) = ln0

DNE or

cos x + sin x = 0sin x = −cos xsin xcos x

= −cos xcos x

tan x = −1

x =− π

4± 2πn

3π4

± 2πn= π

4±πn⎧

⎨⎩

⎧

⎨⎪⎪

⎩⎪⎪

c.v. : x = π4

, 3π4

, −π , π

−π , 0( ), π , 0( ),

− π4

, − 0.322⎛⎝⎜

⎞⎠⎟ , 3π

4, 7.460⎛

⎝⎜⎞⎠⎟

Range: y∈ 0, 1[ ]

11-4 Multiple Choice Homework

1. C

′f x( ) = 2sin x ⋅cos x > x = ′g x( )

−π −π/2 π/2 π 3π/2

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

x

y

20

3. E I. limx→0

sin xx

= 00

=LHlimx→0

cos x1

= 1≠ ∞

Since the limit is finite (not infinite) we have a POE rather than a VA at x = 0 . II. limx→±∞

sin xx

= 0

Since the limit is 0 (which is finite) we have an HA of y = 0 . III. sin xx

= 0

⇒ sin x = 0

⇒ x = 0 ± 2πnπ ± 2πn

⎧⎨⎩

= 0 ±πn{

We have an infinite number of zeros. Statements II and III are true. 5. B dydx

= 2cos 2x( ) ⋅−sin 2x( ) ⋅2= −4cos2xsin2x

336

General Trigonometric Functions Practice Test Answer Key

Multiple Choice – Calculator Allowed

1. E ′f x( ) = cos e− x( ) ⋅e− x ⋅ −1( )

= −e− x cos e− x( )

2. B ′y = 1− sin x′y x=0 = 1− sin0 = 1

y −1= 1 x − 0( )⇒ y = x +1

3. B

Graph ′f x( ) = cos2 xx

− 15

.

The sign of the derivative changes 3 times – i.e. from + to –, then – to +, and finally + to – again.

672

4. D

′E B( ) = 2 B +1( ) ⋅ dBdt

= 2 B +1( ) ⋅20cos t10

⎛⎝⎜

⎞⎠⎟ ⋅

110

= 2 20sin t10

⎛⎝⎜

⎞⎠⎟ + 50 +1⎡

⎣⎢⎤⎦⎥⋅20cos t

10⎛⎝⎜

⎞⎠⎟ ⋅

110

′E 100( ) = 4 20sin 10010

⎛⎝⎜

⎞⎠⎟ + 51⎛

⎝⎜⎞⎠⎟

cos 10010

⎛⎝⎜

⎞⎠⎟ = −134.396 dollars/day

Free Response – Calculator Allowed

1. No fractions with a denominator of zero, no radicals, and no logarithms ∴ Domain: x ∈ −2π , 2π[ ]

′y = 32

− sin x = 0, DNE , EoaASD

sin x = 32

x =

π3± 2πn

2π3

± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

c.v. : x = − 5π3

, − 4π3

, π3

, 2π3

, − 2π , 2π

′yx −2π − 5π

3 4π

3 π

3 2π

3 2π

← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ END + 0 − 0 + 0 − 0 + END

Extreme Points:

−2π , − 4.441( ), 2π , 6.441( ), π

3 , 1.407⎛⎝⎜

⎞⎠⎟, 2π

3 , 1.314⎛⎝⎜

⎞⎠⎟

,

−5π

3 , − 4.034⎛⎝⎜

⎞⎠⎟, − 4π

3 , − 4.128⎛⎝⎜

⎞⎠⎟

673

2. No fractions with a denominator of zero, no radicals, but there is a logarithm –

tan xx 0 π

2 π 3π

2 2π

← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯0 + DNE − 0 + DNE − 0

∴ Domain: x ∈ 0, π2

⎛⎝⎜

⎞⎠⎟ ∪ π , 3π

2⎛⎝⎜

⎞⎠⎟

′y = 1

tan x⋅sec2 x

= 1sin xcos x

= 0 , DNE, EoaASD

If ′y = DNE , then

sin x = 0

x = 0 ± 2πnπ ± 2πn

= 0 ±πn{⎧⎨⎩

or

cos x = 0

x =

π2± 2πn

− π2± 2πn

= ± π2± 2πn⎧

⎨⎩

⎧

⎨⎪⎪

⎩⎪⎪

c.v. : x = 0 , π , 2π , π2

, 3π2

No critical values No extreme points

674

Multiple Choice – No Calculator Allowed

5. E ′f x( ) = sec2 2x( ) ⋅2

′fπ6

⎛⎝⎜

⎞⎠⎟ = sec

2 π3

⎛⎝⎜

⎞⎠⎟ ⋅2 = 8

6. D ddx

sin xy( ) = x⎡⎣ ⎤⎦

cos xy( ) x ⋅ dydx

+ y ⋅1⎡⎣⎢

⎤⎦⎥= 1

xcos xy( ) dydx

+ ycos xy( ) = 1

xcos xy( ) dydx

= 1− ycos xy( )dydx

=1− ycos xy( )xcos xy( )

7. A y x=0 = arctan0 = 0

′y = 1

1+ x3

⎛⎝⎜

⎞⎠⎟2 ⋅13

′yx=0 = 11+ 0( )2

⋅ 13= 13

Point: 0, 0( )

Slope: 13

y − 0 = 13x − 0( )

y = 13x⇔ 3y = x⇔ x − 3y = 0

675

8. B

′f x( ) = limh→0

f x + h( )− f x( )h

= limh→0

cos x + h( )− cos xh

⇒ f x( ) = cos x∴ ′f x( ) = −sin x

OR

′f x( ) = limh→0

cos x + h( )− cos xh

= 00

=LHlimh→0

−sin x + h( ) ⋅1− 01

= limh→0

−sin x + h( )1

= −sin x

Note: When using L’ Hopital’s Rule, you must take the derivative with respect to h.

676

Free Response – No Calculator Allowed

3. Domain: x∈ −2π , 2π⎡⎣ ⎤⎦

Range: y ∈ −4.441, 6.441⎡⎣ ⎤⎦

y-int: 0, 1( ) Axis Points: cos x = 0

x =

π2± 2πn

− π2± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

= ± π2± 2πn⎧

⎨⎩

⇒ x = − 3π2

, − π2

, π2

, 3π2

π2 , 3π

4⎛

⎝⎜

⎞

⎠⎟ , π

2 , − 3π4

⎛

⎝⎜

⎞

⎠⎟ , 3π

2 , 3 3π4

⎛

⎝⎜

⎞

⎠⎟ , − 3π

2 , − 3 3π4

⎛

⎝⎜

⎞

⎠⎟

Extreme Points: −2π , − 4.441( ), 2π , 6.441( ), π

3 , 1.407⎛⎝⎜

⎞⎠⎟, 2π

3 , 1.314⎛⎝⎜

⎞⎠⎟

,

−5π

3 , − 4.034⎛⎝⎜

⎞⎠⎟, − 4π

3 , − 4.128⎛⎝⎜

⎞⎠⎟

VAs: None POEs: None End Behavior: None

677

4. Domain: x ∈ 0, π

2⎛⎝⎜

⎞⎠⎟∪ π , 3π

2⎛⎝⎜

⎞⎠⎟

Range: y∈All Reals y-int: None Zeros:

ln tan x( ) = 0

eln tan x( ) = e0

tan x = 1

x =

π4± 2πn

5π4

± 2πn

⎧

⎨⎪⎪

⎩⎪⎪

= π4±πn⎧

⎨⎩

⇒ x = π4

, 5π4

π4

, 0⎛⎝⎜

⎞⎠⎟ , 5π

4, 0⎛

⎝⎜⎞⎠⎟

Extreme Points: None

VAs: x = 0, x = π

2, x = π , x = 3π

2

POEs: None End Behavior: None