AF12 Chapter 3 Solutions PDF

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Chapter 3

Rational Functions

Chapter 3 Prerequisite Skills Chapter 3 Prerequisite Skills

Question 1 Page 146

Answers may vary. A sample solution is shown. A line or curve that the graph approaches more and more closely. For f(x) =

1 , the vertical x

asymptote is x = 0. Chapter 3 Prerequisite Skills

Question 2 Page 146

a) x = 3 , y = 0

b) x = –4, y = 0

c) x = 8, y = 0

MHR • Advanced Functions 12 Solutions 246

d) x = 5, y = 0

Chapter 3 Prerequisite Skills

Question 3 Page 146

a)

No restrictions on the domain or range. domain: {x ∈ R }, range: {y ∈ R } b)

No restrictions on the domain. domain: {x ∈ R } From the graph: range: {y ∈ R , y ≥ 4}

MHR • Advanced Functions 12 Solutions 247

c)

No restrictions on the domain and range. domain: {x ∈ R }, range: {y ∈ R } d)

Division by zero is not defined. x≠0 domain: {x ∈ R , x ≠ 0} From the graph: range: {y ∈ R , y ≠ 0} e)

Division by zero is not defined. x–4≠0 x≠4 domain: {x ∈ R , x ≠ 4} From the graph: range: {y ∈ R , y ≠ 0}

MHR • Advanced Functions 12 Solutions 248

f)

Division by zero is not defined. x≠0 domain: {x ∈ R , x ≠ 0} From the graph: range: {y ∈ R , y ≠ 0} Chapter 3 Prerequisite Skills

Question 4 Page 146

a)

8 + 5 13 = !13 = 2 ! 3 !1

b)

3! 0 3 1 = = 5+4 9 3

c)

2 2 4!2 =! = 9 !7 ! 2 !9

d)

8!9 = !1 1! 0

e)

!6 ! 7 = !13 2 !1

f)

!9 ! 3 !12 6 = = !7 ! 3 !10 5

Chapter 3 Prerequisite Skills a)

c)

e)

7 ! 10 !3 = !1 ! 7 !8 =& 0.38 4!2 2 = 7 + 4 11 =& 0.18 !0.9 + 5.2 4.3 = 1.5 + 6.6 8.1 =& 0.53

Question 5 Page 146 b)

d)

f)

11 ! 6 5 = 7!0 7 =& 0.71 5 4 +1 = 11 + 2 13 =& 0.38 !1.7 + 3.2 1.5 = 10.1 ! 5.8 4.3 =& 0.35

MHR • Advanced Functions 12 Solutions 249

Chapter 3 Prerequisite Skills a) (x + 4)(x + 3)

Question 6 Page 146 b) (5x – 2)(x – 3)

c) (3x + 8)(2x – 1)

d) (x + 1)(x + 3)(x – 2)

" 1% e) P $ ! ' = 0 # 2& (2x + 1) is a factor

6x 2 ! x ! 2 2 x + 1 12x 3 + 4x 2 ! 5 x ! 2 12x 3 + 6x 2 ! 2x2 ! 5x !2x 2 ! x ! 4x ! 2 !4x ! 2 0 (2 x + 1)(6 x 2 ! x ! 2) = (2 x + 1)(2 x + 1)(3x ! 2) = (2 x + 1) 2 (3x ! 2) f) (3x ! 4)(9 x 2 + 12 x + 16) Chapter 3 Prerequisite Skills a) (x ! 8)(x + 4) = 0 x = 8 or x = !4

Question 7 Page 146 b) (x + 5)(x + 1) = 0 x = !5 or x = !1

c) (x ! 3)(2x ! 3) = 0 3 x = 3 or x = 2

d) (x + 5)(6x + 1) = 0

e) (x + 7)(2x ! 1) = 0 1 x = !7 or x = 2

f) (x ! 6)(3x + 5) = 0

x = !5 or x = !

x = 6 or x = !

1 6

5 3

MHR • Advanced Functions 12 Solutions 250

Chapter 3 Prerequisite Skills

b) 2x 2 + 8x + 1 = 0

a) x 2 ! 4x + 2 = 0

x=

4 ± (!4)2 ! 4(1)(2) 2(1)

4± 8 2 x=2± 2 x=

c) !3x 2 + 5x + 4 = 0

x=

!5 ± 5 2 ! 4( !3)(4) 2( !3)

x=

!5 ± 73 !6

x=

5 ± 73 6

e) 3x 2 + 8x + 2 = 0

x=

Question 8 Page 146

!8 ± 8 2 ! 4(3)(2) 2(3)

x=

!8 ± 40 6

x=

!4 ± 10 3

x=

!8 ± 8 2 ! 4(2)(1) 2(2)

!8 ± 56 4 !4 ± 14 x= 2

x=

d) no real roots; no x-intercepts

2 f) !x + 2x + 7 = 0

x=

!2 ± 2 2 ! 4( !1)(7) 2( !1)

!2 ± 32 !2 x =1± 2 2 x=

MHR • Advanced Functions 12 Solutions 251

Chapter 3 Prerequisite Skills

Question 9 Page 147

a) 2x > 12 x>6

b) 9 + 2 ≥ 6x – 4x 11 ≥ 2x 11 x≤ 2

c) 4x – 8x < –2 –4x < –2 1 x> 2

d) 2x – x > –4 – 1 x > –5

e) 3x – x > –1 – 4 2x > –5 5 x> ! 2

MHR • Advanced Functions 12 Solutions 252

f) x – 2x < 2 + 7 –x < 9 x > –9

Chapter 3 Prerequisite Skills

Question 10 Page 147

a) (x – 2)(x + 2) ≤ 0 Case 1 x≤2 x ≥ –2 –2 ≤ x ≤ 2 is a solution.

–5

–4

–3

–2

–1

0

1

2

3

4

5

Case 2 x≥2 x ≤ –2 No solution. The solution is –2 ≤ x ≤ 2. b) (x – 6)(x + 3) > 0 Case 1 x>6 x > –3 x > 6 is a solution.

–5

–4

–3

–2

–1

0

1

2

3

4

5

Case 2 x 6.

MHR • Advanced Functions 12 Solutions 253

c)

2x2 – 26 = 0 2(x2 – 13) = 0 x2 = 13 x = 13 or x = ! 13

(

)(

–5

–4

–3

–2

–1

0

1

2

3

4

5

)

2 x ! 13 x + 13 < 0 Case 1 x < 13 x > ! 13 ! 13 < x < 13 is a solution. Case 2 x > 13 No solution.

x < ! 13

The solution is ! 13 < x < 13 . d) 3x2 – 2x2 + 5 x – 2 x – 12 + 2 > 0 x2 + 3x – 10 > 0 (x + 5)(x – 2) > 0 –7

–6

–5

–4

–3

–2

–1

0

1

2

3

Case 1 x > –5 x>2 x > 2 is a solution. Case 2 x < –5 x 2.

MHR • Advanced Functions 12 Solutions 254

e) 2x2 – x2 – x + 9x + 4 + 3 < 0 x2 + 8x + 7 < 0 (x + 7)(x + 1) < 0 –7

–6

–5

–4

–3

–2

–1

0

1

2

3

Case 1 x < –7 x > –1 No solution. Case 2 x > –7 x < –1 –7 < x < –1 is a solution. The solution is –7 < x < –1. f) x2 + x2 + 2 x + 9 x + 2 – 8 > 0 2x2 + 11 x – 6 > 0 (x + 6)(2x – 1) > 0 –7

Case 1 x > –6 x>

x>

–6

–5

–4

–3

–2

–1

0

1

2

3

1 2

1 is a solution. 2

Case 2

1 2 x < –6 is a solution. x < –6

x<

1 The solution is x < –6 or x > . 2

MHR • Advanced Functions 12 Solutions 255

Chapter 3 Section 1

Reciprocal of a Linear Function

Chapter 3 Section 1

Question 1 Page 153

a) As x → 2+ 2– +∞ ∞ –∞ ∞

f(x) → +∞ ∞ –∞ ∞ 0 0

As x → –5+ –5– +∞ ∞ –∞ ∞

f(x) → +∞ ∞ –∞ ∞ 0 0

As x → 8+ 8– +∞ ∞ –∞ ∞

f(x) → +∞ ∞ –∞ ∞ 0 0

b)

c)

Chapter 3 Section 1

Question 2 Page 153

a) i) x = 2, y = 0 ii) x = –3, y = 0

1 shifted 2 to the right 2x 1 y= 2( x ! 2)

b) i) y =

1 shifted 3 to the left 2x 1 y= 2( x + 3)

ii) y =

MHR • Advanced Functions 12 Solutions 256

Chapter 3 Section 1 a) i) x = 5

Question 3 Page 154 ii) y = 0

iii) let x = 0

1 0!5 1 =! 5

f (0) =

b) i) x = –6

ii) y = 0

iii) let x = 0

2 0+6 1 = 3

g(0) =

c) i) x = 1

ii) y = 0

iii) let x = 0

5 1! 0 =5

h(0) =

d) i) x = –7

ii) y = 0

iii) let x = 0

1 0+7 1 =! 7

k(0) = !

Chapter 3 Section 1

Question 4 Page 154

a)

b)

MHR • Advanced Functions 12 Solutions 257

c)

d)

Chapter 3 Section 1

Question 5 Page 154

a) Since the vertical asymptote is x = 3: 1 y= x!3 b) Since the vertical asymptote is x = –3: 1 y= x+3 c) Since the vertical asymptote is x =

y=

1 : 2

1 2x ! 1

d) Since the vertical asymptote is x = –4 and it is reflected in the y-axis: 1 y=! x+4

MHR • Advanced Functions 12 Solutions 258

Chapter 3 Section 1

Question 6 Page 154

a)

Select a few points to the left of the asymptote and analyse the slope. At x = –1, f(x) = –0.25 At x = 0, f(x) =& –0.33 !0.33 + 0.25 Slope = 0 +1 = –0.08 At x = 1, f(x) = –0.5 At x = 2, f(x) = –1 !1 + 0.5 Slope = 2 !1 = –0.5 Since –0.5 < –0.08, the slope is negative and decreasing for the interval x < 3. Select a few points to the right of the asymptote and analyse the slope. At x = 3.5, f(x) = 2 At x = 4, f(x) = 1 1! 2 Slope = 4 ! 3.5 = –2 At x = 5, f(x) = 0.5 At x = 6, f(x) =& 0.33 0.33 ! 0.5 Slope = 6!5 = –0.17 Since –0.17 > –2, the slope is negative and increasing for the interval x > 3.

MHR • Advanced Functions 12 Solutions 259

b)

Select a few points to the left of the asymptote and analyse the slope. At x = –6, f(x) = –0.6 At x = –5, f(x) = –1 !1 + 0.6 Slope = !5 + 6 = –0.4 At x = –4, f(x) = –3 At x = –3.8, f(x) = –5 !5 + 3 Slope = !3 + 4 = –2

7 Since –2 < –0.4, the slope is negative and decreasing within the interval x < ! . 2 Select a few points to the right of the asymptote and analyse the slope. At x = –3, f(x) = 3 At x = –2, f(x) = 1 1! 3 Slope = !2 + 3 = –2 At x = –1, f(x) = 0.6 At x = 0, f(x) =& 0.43 0.43 ! 0.6 Slope = 0 +1 = –0.17 Since –0.17 > –2, the slope is negative and increasing for x >!

7 . 2

MHR • Advanced Functions 12 Solutions 260

c)

Select a few points to the left of the asymptote and analyse the slope. At x = –5, f(x) = 2 At x = –4.5, f(x) = 4 4!2 Slope = !4.5 + 5 =4 Since 4 > 0.33, the slope is positive and increasing for x < –4. Select a few points to the right of the asymptote and analyse the slope. At x = –3, f(x) = –2 At x = –2, f(x) = –1 !1 + 2 Slope = !2 + 3 =1 At x = 0, f(x) = –0.5 At x = 1, f(x) = –0.4 !0.4 + 0.5 Slope = 1! 0 = 0.1 Since 0.1 < 1, the slope is positive and decreasing for x > –4.

MHR • Advanced Functions 12 Solutions 261

d)

Select a few points to the left of the asymptote and analyse the slope. At x = –2, f(x) =& 0.71 At x = –1, f(x) = 1 1 ! 0.71 Slope =& !1 + 2 =& 0.29 At x = 0, f(x) =& 1.67 At x= 1, f(x) = 5 5 ! 1.67 Slope =& 1! 0 =& 3.33

3 Since 3.33 > 0.29, the slope is positive and increasing for x < . 2 Select a few points to the right of the asymptote and analyse the slope. At x = 2, f(x) = –5 At x = 3, f(x) =& –1.67 !1.67 + 5 Slope =& 3! 2 =& 3.33 At x = 4, f(x) = –1 At x = 5, f(x) =& –0.71 !0.71 + 1 Slope =& 5!4 =& 0.83

3 Since 0.83 < 3.33, the slope is positive and decreasing for x > . 2

MHR • Advanced Functions 12 Solutions 262

Chapter 3 Section 1

Question 7 Page 154

a)

{x ∈ R , x ≠ 1}, {y ∈ R , y ≠ 0}, x = 1, y = 0 b)

{x ∈ R , x ≠ –4}, {y ∈ R , y ≠ 0}, x = –4, y = 0 c)

{x ∈ R , x ≠ –

1 1 }, {y ∈ R , y ≠ 0}, x = – , y = 0 2 2

d)

{x ∈ R , x ≠ –4}, {y ∈ R , y ≠ 0}, x = –4, y = 0

MHR • Advanced Functions 12 Solutions 263

e)

{x ∈ R , x ≠

5 5 }, {y ∈ R , y ≠ 0}, x = , y = 0 2 2

f)

{x ∈ R , x ≠ 5}, {y ∈ R , y ≠ 0}, x = 5, y = 0 g)

{x ∈ R , x ≠

1 1 }, {y ∈ R , y ≠ 0}, x = , y = 0 4 4

h)

{x ∈ R , x ≠ –

1 1 }, {y ∈ R , y ≠ 0}, x = – , y = 0 2 2

MHR • Advanced Functions 12 Solutions 264

Chapter 3 Section 1

Question 8 Page 154

For the y-intercept let x = 0. 1 = !1 f (0) = 0!c 1 ! = !1 c c =1 1 f (x) = kx ! 1 For the asymptote let x = 1. kx ! 1 = 0 kx = 1

k(1) = 1 k =1 1 f (x) = x !1 Chapter 3 Section 1

Question 9 Page 154

Let x = 0. 1 f (0) = 0!c = !0.25

!

1 = !0.25 c c=4

f (x) =

1 kx ! 4

Asymptote is at x = –1: kx ! 4 = 0 kx = 4

k(!1) = 4 k = !4 1 !4x ! 4 1 f (x) = ! 4x + 4 1 y=! 4x + 4 f (x) =

MHR • Advanced Functions 12 Solutions 265

Chapter 3 Section 1

Question 10 Page 155

a) d = 350 × 11 = 3850 3850 t= v b)

3850 500 t = 7.7

c) t =

It would take 7.7 h or 7 h and 42 min. d) As the speed increases the rate of change of time decreases. Chapter 3 Section 1

Question 11 Page 155

a) Answers may vary. b) Answers may vary. A sample solution is shown. 2 The equation of the asymptote is x = ! . b When b = 1, the asymptote is x = –2. 2 When b > 1, !2 < ! < 0 , the vertical asymptote is between –2 and 0. b 2 When 0 < b < 1, ! < !2 , the vertical asymptote is less than –2. b When b < 0, the vertical asymptote is bigger than zero. Chapter 3 Section 1

Question 12 Page 155

a)

MHR • Advanced Functions 12 Solutions 266

b)

c)

Chapter 3 Section 1 a)

Question 13 Page 155

f = 200 × 3 600 F= d

b)

600 2 F = 300 N

c) F =

300 N of force is needed to lift the object 2 m from the fulcrum.

600 2d 300 F= d

d) F =

The force is halved.

MHR • Advanced Functions 12 Solutions 267

Chapter 3 Section 1

Question 14 Page 155

a) Since division by zero and negative square roots are not defined, x > 0 and y > 0. domain: {x ∈ R , x > 0} range: {y ∈ R , y > 0} vertical asymptote: x = 0 horizontal asymptote: y = 0

b) Since division by 0 is not defined, x ≠ 0 and y ≠ 0. Since x is an absolute value, g(x) is positive. domain: {x ∈ R , x ≠ 0} range: {y ∈ R , y > 0} vertical asymptote: x = 0 horizontal asymptote: y = 0

MHR • Advanced Functions 12 Solutions 268

c) Division by zero is not defined. x!2"0 x"2 domain: {x ∈ R , x ≠ 2} To find the range, first find the inverse function. 3 +4 x= y!2 3 x!4= y!2

3 x!4 3 +2 y= x!4 Division by zero is not defined. x!4"0 y!2=

x"4 The domain of the inverse function is the range of f(x). range: {y ∈ R , y ≠ 4} vertical asymptote: x = 2 horizontal asymptote: y = 4

MHR • Advanced Functions 12 Solutions 269

Chapter 3 Section 1

Question 15 Page 155

Left side of the x-intercept:

Coordinates for f (x) =

At x = 2, y = –1, reciprocal = –1 1 At x = 1, y = –3, reciprocal = ! 3 1 At x = 0, y = –5, reciprocal= ! 5 1 At x = –1, y = –7, reciprocal = ! 7

(2, –1) 1% " $#1,!! '& 3

Right side of the x-intercept: At x = 3, y = 1, reciprocal = 1 1 At x = 4, y = 3, reciprocal = 3

1 5 1 At x = 6, y = 7, reciprocal = 7 At x = 5, y = 5, reciprocal =

1 2x ! 5

1% " $# 0,!! '& 5 1% " $# !1,!! '& 7

(3, 1) ! 1$ #" 4,! &% 3

! 1$ #" 5,! &% 5 ! 1$ #" 6,! &% 7

Answers may vary. A sample solution is shown. The reciprocal of the y-coordinates on either side of the x-intercept (y = 2x – 5) are the 1 y-coordinates of f (x) = . 2x ! 5

MHR • Advanced Functions 12 Solutions 270

Chapter 3 Section 1

Question 16 Page 155

1 1 1 = ! x z y 1 y z = ! x zy zy 1 y!z = x zy zy x= y!z x=

yz ,!y " z,!x " 0,!z " 0 y!z

Chapter 3 Section 1

Question 17 Page 155

75b ! 5b 70b = 5b 5b !!!!!!!!!!!!!!!= 14 Chapter 3 Section 1 E

Question 18 Page 155

2 3 A 1

1 B

If two points are within 1 unit of each other, the angle between them must be less than

! 3

(they form an equilateral triangle). That means that given any point A, if point B is in the nearest third of the circle to A, the distance 2 of the circle where the distance is greater than 1 unit. will be less than 1 unit. So there is 3

MHR • Advanced Functions 12 Solutions 271

Chapter 3 Section 2

Reciprocal of a Quadratic Function

Chapter 3 Section 2

Question 1 Page 164

a) As x → 3– 3+ 1– 1+ –∞ +∞

f(x) → –∞ +∞ +∞ –∞ 0 0

As x → –4– –4+ 5– 5+ –∞ +∞

f(x) → +∞ –∞ –∞ +∞ 0 0

As x → –6– –6+ –∞ +∞

f(x) → –∞ –∞ 0 0

b)

c)

Chapter 3 Section 2

Question 2 Page 165

a) asymptote: x = 4

domain: {x ∈ R , x ≠ 4}

b) asymptotes: x = 2, x = –7

domain: {x ∈ R , x ≠ 2, x ≠ –7}

c) No asymptotes or restrictions on the domain. domain: {x ∈ R }

3 (x ! 5)( x + 5) asymptotes: x = –5, x = 5

d) m(x) =

domain: {x ∈ R , x ≠ 5, x ≠ –5}

MHR • Advanced Functions 12 Solutions 272

1 (x ! 3)( x ! 1) asymptotes: x = 3, x = 1

domain: {x ∈ R , x ≠ 1, x ≠ 3}

2 (x + 4)( x + 3) asymptotes: x = -4, x = –3

domain: {x ∈ R , x ≠ –4, x ≠ –3}

e) h(x) =

f) k(x) = !

g) n(x) = !

2 (x + 2)(3x ! 4)

asymptotes: x = –2, x =

4 3

domain: {x ∈ R , x ≠ –2, x ≠

4 } 3

h) No asymptotes or restrictions on the domain. domain: {x ∈ R } Chapter 3 Section 2

Question 3 Page 165

a) Interval x1

Sign of f(x) + +

Sign of Slope + –

Change in Slope + –

Interval x < –2 –2 < x < 1 x=1 15

Sign of f(x) + – – – +

Sign of Slope + + 0 – –

Change in Slope + – – – +

vi) {y ∈ R , y ≠ 0}

MHR • Advanced Functions 12 Solutions 276

c) i) The denominator cannot equal zero, there are restrictions at x 2 + 5x ! 21 = 0

x=

!5 ± 5 2 ! 4(1)(!21) 2(1)

!5 + 109 !5 ! 109 or x = 2 2 $& #5 ± 109 (& domain: % x !R,!x " ) 2 *& '& x=

!5 + 109 !5 ! 109 ,!x = 2 2 As x → ±∞, the denominator approaches +∞, so f(x) approaches 0. horizontal asymptote: y = 0

ii) vertical asymptotes at x =

iii) let x = 0

p(0) = !

1 2

0 + 5(0) ! 21 1 = 21 1 y-intercept: 21 iv)

MHR • Advanced Functions 12 Solutions 277

v) Interval !5 ! 109 x< 2 !5 ! 109 < x < –2.5 2 x = –2.5 !5 + 109 –2.5 < x < 2 !5 + 109 x> 2

Sign of f(x)

Sign of Slope

Change in Slope

–

–

–

+

–

+

+

0

+

+

+

+

–

+

–

vi) {y ∈ R , y ≠ 0} d) i) w(x) =

1 (x ! 2)(3x + 1)

1' $ domain: % x !R,! x " 2,! x " # ( 3) &

1 3 As x → ±∞, the denominator approaches +∞, so f(x) approaches 0. horizontal asymptote: y = 0

ii) vertical asymptotes: x = 2, x = !

iii) let x = 0

1 (0 ! 2)(3(0) + 1) 1 =! 2 1 y-intercept: ! 2 w(0) =

iv)

MHR • Advanced Functions 12 Solutions 278

v) Interval 1 x< ! 3 5 1 ! 0: the function is positive and decreasing (negative slope) b) domain and range: {x !R,!x " 1} , {y !R,!y > 0 } asymptotes: x = 1, y = 0 y-intercept: 1 x < 1: the function is positive and increasing (positive slope) x > 1: the function is positive and decreasing (negati...