and Chapter
4
Polynomial Rational Functions
4.3
1. 3.
Rational Functions I
True y= 1 x 2. 4. Quotient: 3x - 6; Remainder: 6x -10 True
5. 7. 9. 11.
y =1 proper True
6. 8. 10.
x = -1 False True
4x In R(x) = x - 3 , the denominator, q(x ) = x - 3, has a zero at 3. Thus, the domain of R(x) is { x x 3} . 5x 2 In R(x) = 3 + x , the denominator, q(x ) = 3 + x , has a zero at 3. Thus, the domain of R(x) is { x x -3} . -4x2 , the denominator, q(x ) = (x - 2)(x + 4), has zeros at 2 and 4. (x - 2)(x + 4) Thus, the domain of H(x) is { x x -4, x 2} . In H(x) = 6 , the denominator, q(x ) = (x + 3)(4 - x) , has zeros at 3 and 4. (x + 3)(4 - x) Thus, the domain of G(x) is { x x -3, x 4} . In G(x) = 446
12.
13.
14.
Section 4.3
15. In F(x) =
Rational Functions I
3x(x - 1) , the denominator, q(x ) = 2x 2 - 5x - 3 = (2x + 1)(x - 3), has zeros 2x 2 - 5x - 3 1 1 at - and 3 . Thus, the domain of F(x) is x x - , x 3 . 2 2 -x(1- x) , the denominator, q(x ) = 3x 2 + 5x - 2 = (3x - 1)(x + 2), has zeros 2 3x + 5x - 2 1 1 at and - 2. Thus, the domain of Q(x) is x x -2,x . 3 3 In Q(x) = x , the denominator, q(x ) = x 3 - 8 = (x - 2)(x 2 + 2x + 4) , has a zero at 2. x3 - 8 ( x 2 + 2x + 4 has no real zeros.) Thus, the domain of R(x) is { x x 2} . In R(x) = x , the denominator, q(x ) = x 4 - 1 = (x - 1)(x + 1)(x 2 + 1), has zeros at 1 x -1 and 1. ( x 2 + 1 has no real zeros.) Thus, the domain of R(x) is { x x -1,x 1} . In R(x) =
4
16.
17.
18.
19.
In H(x) =
3x 2 + x , the denominator, q(x ) = x 2 + 4 , has no real zeros. Thus, the domain of 2 x +4 H(x) is { x x is a real number} .
20.
In G(x) =
x-3 , the denominator, q(x ) = x 4 + 1, has no real zeros. Thus, the domain of x4 + 1 G(x) is { x x is a real number} .
21.
3(x 2 - x - 6 ) , the denominator, q(x ) = 4(x 2 - 9) = 4(x - 3)(x + 3) , has zeros at 4(x 2 - 9) 3 and 3. Thus, the domain of R(x) is { x x -3,x 3} . In R(x) = - 2(x 2 - 4) , the denominator, q(x ) = 3(x 2 + 4x + 4) = 3(x + 2) 2 , has a zero 2 3(x + 4x + 4) at 2. Thus, the domain of F(x) is { x x -2} . In F(x) = (a) (b) (d) (a) (b) (d) Domain: {x x 2} ; Range: Intercept: (0, 0) Vertical Asymptote: x = 2 Domain: {x x -1} ; Range: Intercept: (0, 2) Vertical Asymptote: x = -1
22.
23.
{y
y 1} (c) (e) y > 0} (c) (e)
Horizontal Asymptote: y = 1 Oblique Asymptote: none
24.
{y
Horizontal Asymptote: y = 0 Oblique Asymptote: none
447
Chapter 4
25. (a) (b) (d) (a) (b) (d) (a) (b) (d) (a) (b) (d)
Polynomial and Rational Functions
Domain: {x x 0} ; Range: all real numbers Intercepts: (1, 0), (1, 0) (c) Horizontal Asymptote: none Vertical Asymptote: x = 0 (e) Oblique Asymptote: y = 2x Domain: {x x 0} ; Range: Intercepts: none Vertical Asymptote: x = 0
26.
{y
y 2 or y -2 } (c) Horizontal Asymptote: none (e) Oblique Asymptote: y = - x
27.
Domain: {x x - 2, x 2} ; Range: {y y 0 or y > 1} Intercept: (0, 0) (c) Horizontal Asymptote: y = 1 Vertical Asymptotes: x = - 2, x = 2 (e) Oblique Asymptote: none Domain: {x x -1 x 1} ; Range: all real numbers , Intercept: (0, 0) (c) Horizontal Asymptote: y = 0 Vertical Asymptotes: x = - 1 x = 1 (e) Oblique Asymptote: none , 1 x 1 , shift the graph x 30. Q(x) = 3 + 1 x2
28.
29.
F ( x) = 2 +
Using the function, y = vertically 2 units up.
1 , shift the x2 graph vertically 3 units up. Using the function y =
31.
R( x) =
1 2 ( x - 1)
32.
Q(x) =
1 3 = 3 x x
1 , shift the x2 graph horizontally 1 unit to the right. Using the function, y =
1 Using the function y = , stretch the x graph vertically by a factor of 3.
448
Section 4.3
33. -2 1 = -2 x +1 x +1 Using the function y = 1 , shift the graph x horizontally 1 unit to the left, reflect about the x-axis, and stretch vertically by a factor of 2. H(x) = 34. G(x) =
Rational Functions I
2 1 2 = 2 (x + 2)2 (x + 2) 1 Using the function y = 2 , shift the x graph horizontally 2 units to the left, and stretch vertically by a factor of 2.
35.
-1 1 36. =- 2 x + 4x + 4 ( x + 2) 1 Using the function y = 2 , shift the graph x horizontally 2 units to the left, and reflect about the x-axis. R(x) =
2
1 R(x) = x - 1 + 1 1 Using the function y = , shift the graph x horizontally 1 unit to the right, and shift vertically 1 unit up.
37.
2 2 +1 2 = (x - 3) (x - 3)2 1 + 1 = 2 (x - 3)2 1 Using the function y = 2 , shift the x graph right 3 units, stretch vertically by a factor of 2, and shift vertically 1 unit up. G(x) =1+
38.
1 1 = - +2 x +1 x +1 1 Using the function y = x , shift the graph left 1 unit, reflect about the x-axis, and shift vertically up 2 units. F(x) = 2 -
449
Chapter 4
Polynomial and Rational Functions
1 1 x- 4 4 x2 - 4 4 R(x) = =1- = -4 +1 =1- 2 = -4 2 +1 39. 40. x x x x x x2 1 1 Using the function y = x , reflect about Using the function y = 2 , reflect about x the x-axis, stretch vertically by a factor the x-axis, stretch vertically by a factor of of 4, and shift vertically 1 unit up. 4, and shift vertically 1 unit up.
R(x) =
41.
3x R(x) = x + 4 The degree of the numerator, p(x) = 3x, is n = 1. The degree the of denominator, 3 q(x) = x + 4, is m = 1. Since n = m , the line y = 1 = 3 is a horizontal asymptote. The denominator is zero at x = - 4, so x = - 4 is a vertical asymptote. 3x + 5 R(x) = x - 6 The degree of the numerator, p(x) = 3x + 5, is n = 1. The degree of the denominator, 3 q(x ) = x - 6, ism = 1. Since n = m , the line y = 1 = 3 is a horizontal asymptote. The denominator is zero at x = 6 , so x = 6 is a vertical asymptote. x 4 + 2x 2 + 1 x2 - x + 1 The degree of the numerator, p(x) = x 4 + 2x 2 + 1, is n = 4 . The degree of the denominator, q(x ) = x 2 - x + 1, is m = 2. Since n > m + 1, there is no horizontal asymptote or oblique asymptote. The denominator has no real zeros, so there is no vertical asymptote. H(x) =
42.
43.
450
Section 4.3
44.
Rational Functions I
-x 2 + 1 G(x) = x + 5 The degree of the numerator, p(x) = -x 2 + 1, is n = 2. The degree of the denominator, q(x) = x + 5, is m = 1. Since n = m + 1, there is an oblique asymptote. Dividing: -x + 5 x + 5 -x 2 + -x 2 - 5x
)
1
- 24 G(x) = -x + 5 + x + 5
5x + 1 5x + 25 - 24 Thus, the oblique asymptote is y = - x + 5. The denominator is zero at x = -5, so x = -5 is a vertical asymptote. 45. x3 x4 - 1 The degree of the numerator, p(x) = x 3 , is n = 3. The degree of the denominator, q(x ) = x 4 - 1 is m = 4 . Since n < m , the line y = 0 is a horizontal asymptote. The denominator is zero at x = -1 and x = 1, so x = -1 and x = 1 are vertical asymptotes. T (x) = 4x 5 x3 - 1 The degree of the numerator, p(x) = 4x 5 , is n = 5 . The degree of the denominator, q(x ) = x 3 - 1 is m = 3. Since n > m + 1, there is no horizontal asymptote or oblique asymptote. The denominator is zero at x = 1, so x = 1 is a vertical asymptote. P(x) = 5 - x2 3x 4 The degree of the numerator, p(x) = 5 - x 2 , is n = 2 . The degree of the denominator, q(x ) = 3x 4 is m = 4 . Since n < m , the line y = 0 is a horizontal asymptote. The denominator is zero at x = 0 , so x = 0 is a vertical asymptote. Q(x) = - 2x 2 + 1 - 2x 2 + 1 = 2 2x 3 + 4x 2 2x (x + 2) The degree of the numerator, p(x) = - 2x 2 + 1, is n = 2. The degree of the denominator, q(x ) = 2x 3 + 4x 2 is m = 3. Since n < m , the line y = 0 is a horizontal asymptote. The denominator is zero at x = 0 and x = - 2 , so x = 0 and x = - 2 are vertical asymptotes. F(x) =
46.
47.
48.
451
Chapter 4
49. R(x) =
Polynomial and Rational Functions
3x 4 + 4 x 3 + 3x The degree of the numerator, p(x) = 3x 4 + 4, is n = 4 . The degree of the denominator, q(x ) = x 3 + 3x is m = 3 . Since n = m + 1, there is an oblique asymptote. Dividing: 3x 2 3 x + 3x 3x 4 + 4 R(x) = 3x + -9x + 4 x 3 + 3x 3x 4 + 9x 2
)
- 9x 2 +4 Thus, the oblique asymptote is y = 3x . The denominator is zero at x = 0 , so x = 0 is a vertical asymptote. 50. 6x 2 + x + 12 6x 2 + x + 12 = (3x + 1)(x - 2) 3x 2 - 5x - 2 The degree of the numerator, p(x) = 6x 2 + x + 12, is n = 2 . The degree of the denominator, 6 q(x ) = 3x 2 - 5x - 2 is m = 2. Since n = m , the line y = 3 = 2 is a horizontal asymptote. 1 1 The denominator is zero at x = - 3 and x = 2 , so x = - 3 and x = 2 are vertical asymptotes. R(x) = x 3 -1 x - x2 The degree of the numerator, p(x) = x 3 - 1 is n = 3. The degree of the denominator, , 2 q(x ) = x - x is m = 2 . Since n = m + 1, there is an oblique asymptote. Dividing: -x -1 2 3 x -1 1 -x + x x -1 G(x) = -x -1+ = -x -1- , x 1 2 x- x x x3 - x2 Thus, the oblique asymptote is y = -x - 1. x2 G(x) =
51.
)
x2 - x x -1 G(x) must be in lowest terms to find the vertical asymptote: x 3 - 1 (x - 1)(x 2 + x + 1) x 2 + x + 1 G(x) = = = -x(x - 1) -x x - x2 The denominator is zero at x = 0 , so x = 0 is a vertical asymptote. x -1 x -1 x -1 1 = -x(x - 1)(x + 1) = - x(x + 1) 3 = 2 x -x - x(x - 1) The degree of the numerator, p(x) = x - 1, is n = 1. The degree of the denominator, q(x ) = x - x 3 is m = 3. Since n < m , the line y = 0 is a horizontal asymptote. The denominator is zero at x = 0, and x = -1, so x = 0 and x = -1 are vertical asymptotes. F(x) =
52.
452
Section 4.3
53. g( h ) = (a) (b) (c) (d)
Rational Functions I
(6.374 10
3.99 1014
6
g(0) =
(6.374 10
3.99 1014
6
+ h)
2
g( 443) =
(6.374 10
3.99 10
6
3.99 1014
6
+ 0)
2
9.8208 m/s 2 9.8195 m/s 2 9.7936 m/s 2
g(8848) = g( h ) =
(6.374 10
14
3.99 1014
6
+ 443)
2
+ 8848)
2
2
(e)
3.99 1014 = 0, which is impossible. Therefore, there is no height above sea level at which g = 0. In other words, there is no point in the entire universe that is unaffected by the Earth's gravity! 54. P(t) = (a) (b) (d) 50(1+ 0.5t ) 2 + 0.01t 50(1+ 0) 50 P (0) = = = 25 insects 2+0 2 50(1+ 0.5(60)) 1550 5 years = 60 months; P (60) = = 596 insects 2 + 0.01(60) 2.6 50(1+ 0.5t ) 50(0.5t ) P(t) = = 2500 as t 2 + 0.01t 0.01t Thus, y = 2500 is the horizontal asymptote. The area can sustain a maximum population of 2500 insects. Answers will vary.
+ h) Thus, g = 0 is the horizontal asymptote. 3.99 1014 g( h ) = 2 = 0 , to solve this equation would require that (6.374 106 + h)
(6.374 10
3.99 1014 0 as h h2
5558.
453