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Chapter 3
Functions and Their Graphs
3.1 Functions
- Function Domain: {Dad, Colleen, Kaleigh, Marissa} Range: {Jan. 8, Mar. 15, Sept. 17} 2. Function Domain: {Bob, Dave, John, Chuck} Range: {Beth, Diane, Linda, Marcia}
- Not a function 4. Function Domain: {Bob, Dave, John, Chuck} Range: {Diane, Linda, Marcia}
- Function Domain: {2, –3, 4, 1} Range: {6, 9, 10} 6. Function Domain: {–2, –1, 3, 4} Range: {3, 5, 7, 12}
- Function Domain: {1, 2, 3, 4} Range: {3} 8. Function Domain: {0, 1, 2, 3} Range: {–2, 3, 7}
- Not a function 10. Not a function
- Function Domain: {–2, –1, 0, 1} Range: {4, 1, 0}
- Function Domain: {–2, –1, 0, 1} Range: {3, 4, 16}
Chapter 3 Functions and Their Graphs
- f ( x ) = 3 x^2 + 2 x − 4
(a) f (0) = 3(0)^2 + 2(0)− 4 = − 4 (b) f ( 1 ) = 3(1) 2 + 2(1)− 4 = 3 + 2 − 4 = 1 (c) f (−1) = 3(−1)^2 + 2(−1) − 4 = 3 − 2 − 4 = − 3 (d) f (− x ) = 3(− x )^2 + 2(− x ) − 4 = 3 x^2 − 2 x − 4 (e) − f ( x ) = −( 3 x^2 + 2 x − 4) = − 3 x^2 − 2 x + 4 (f) f ( x + 1) = 3( x + 1)^2 + 2( x + 1) − 4 = 3( x^2 + 2 x + 1) + 2 x + 2 − 4 = 3 x^2 + 6 x + 3 + 2 x + 2 − 4 = 3 x^2 + 8 x + 1 (g) f (2 x ) = 3(2 x )^2 + 2(2 x ) − 4 = 12 x^2 + 4 x − 4 (h) f ( x + h ) = 3( x + h )^2 + 2( x + h ) − 4 = 3( x^2 + 2 xh + h^2 ) + 2 x + 2 h − 4 = 3 x^2 + 6 xh + 3 h^2 + 2 x + 2 h − 4
- f ( x ) = − 2 x^2 + x − 1 (a) f (0) = − 2(0)^2 + 0 − 1 = − 1 (b) f ( 1 ) = − 2(1)^2 + 1 − 1 = − 2 (c) f (−1) = − 2(−1)^2 + (−1)− 1 = − 4 (d) f (− x ) = − 2(− x )^2 + (− x ) − 1 = − 2 x^2 − x − 1 (e) − f ( x ) = −(− 2 x^2 + x − 1) = 2 x^2 − x + 1 (f) f ( x + 1) = − 2( x + 1)^2 + ( x + 1)− 1 = − 2( x^2 + 2 x + 1) + x + 1 − 1 = − 2 x^2 − 4 x − 2 + x = − 2 x^2 − 3 x − 2 (g) f (2 x ) = − 2(2 x )^2 + ( 2 x ) − 1 = − 8 x^2 + 2 x − 1 (h) f ( x + h ) = − 2( x + h )^2 + ( x + h ) − 1 = − 2( x^2 + 2 xh + h^2 ) + x + h − 1 = − 2 x^2 − 4 xh − 2 h^2 + x + h − 1
- f ( x ) = x x^2 + 1 (a) f (0) =
(b) f ( 1 ) =
(c) f (−1) =
(−1)^2 + 1
(d) f (− x ) = − x (− x )^2 + 1
− x x^2 + 1 (e) − f ( x ) = − x x^2 + 1
− x x^2 + 1 (f) f ( x + 1) = x + 1 ( x + 1)^2 + 1
x + 1 x^2 + 2 x + 1 + 1
x + 1 x^2 + 2 x + 2 (g) f (2 x ) = 2 x (2 x )^2 + 1
2 x 4 x^2 + 1 (h) f ( x + h ) = x + h ( x + h )^2 + 1
x + h x^2 + 2 xh + h^2 + 1
Chapter 3 Functions and Their Graphs
- f ( x ) = 2 x + 1 3 x − 5 (a) f (0) =
(b) f ( 1 ) =
(c) f (−1) =
(d) f (− x ) = 2(− x ) + 1 3(− x ) − 5
− 2 x + 1 − 3 x − 5
2 x − 1 3 x + 5 (e) − f ( x ) = − 2 x + 1 3 x − 5
− 2 x − 1 3 x − 5 (f) f ( x + 1) = 2( x + 1)+ 1 3( x + 1)− 5
2 x + 2 + 1 3 x + 3 − 5
2 x + 3 3 x − 2 (g) f (2 x ) = 2(2 x ) + 1 3(2 x ) − 5 =^
4 x + 1 6 x − 5 (h) f ( x + h ) = 2( x + h ) + 1 3( x + h ) − 5 =^
2 x + 2 h + 1 3 x + 3 h − 5
- f ( x ) = 1 −
( x + 2)^2 (a) f (0) = 1 −
( 0+ 2 )^2
4 =^
4 (b)^ f^ (^1 )^ =^1 −^
(1+ 2 )^2
9 =^
(c) f (−1) = 1 −
(− 1 + 2)^2
1 =^0
(d) f (− x ) = 1 −
(− x + 2 )^2
x^2 − 4 x + 4
x^2 − 4 x + 4 − 1 x^2 − 4 x + 4
x^2 − 4 x + 3 x^2 − 4 x + 4
(e) − f ( x ) = − 1 −
( x + 2 )^2
x^2 + 4 x + 4
− x^2 − 4 x − 4 + 1 x^2 + 4 x + 4
− x^2 − 4 x − 3 x^2 + 4 x + 4
(f) f ( x + 1) = 1 −
( x + 1 + 2 )^2
( x + 3)^2
x^2 + 6 x + 9 − 1 x^2 + 6 x + 9
x^2 + 6 x + 8 x^2 + 6 x + 9
(g) f (2 x ) = 1 −
(2 x + 2)^2
4 x^2 + 8 x + 4
4 x^2 + 8 x + 4 − 1 4 x^2 + 8 x + 4
4 x^2 + 8 x + 3 4 x^2 + 8 x + 4
(h) f ( x + h ) = 1 −
( x + h + 2 )^2
x^2 + 2 x h + 4 x + h^2 + 4 h + 4 − 1 x^2 + 2 x h + 4 x + h^2 + 4 h + 4
=
x^2 + 2 x h + 4 x + h^2 + 4 h + 3 x^2 + 2 x h + 4 x + h^2 + 4 h + 4
- Graph y = x^2. The graph passes the vertical line test. Thus, the equation represents a function.
- Graph y = x^3. The graph passes the vertical line test. Thus, the equation represents a function.
- Graph y^ =^
x. The graph passes the vertical line test. Thus, the equation represents a function.
Section 3.1 Functions
- Graph y^ =^ x^. The graph passes the vertical line test. Thus, the equation represents a function.
- y^2 = 4 − x^2
Solve for y : y = ± 4 − x^2 For (^) x = 0, y = ± 2. Thus, (0,2) and (0,–2) are on the graph. This is not a function, since a distinct x corresponds to two different y 's.
- (^) y = ± 1 − 2 x
For x = 0, y = ± 1. Thus, (0,1) and (0,–1) are on the graph. This is not a function, since a distinct x corresponds to two different y 's.
- x = y^2
Solve for y : y = ± x For (^) x = 1 , y = ± 1. Thus, (1,1) and (1,–1) are on the graph. This is not a function, since a distinct x corresponds to two different y 's.
- (^) x + y^2 = 1
Solve for (^) y : y = ± 1 − x For (^) x = 0, y = ± 1. Thus, (0,1) and (0,–1) are on the graph. This is not a function, since a distinct x corresponds to two different y 's.
- Graph y = 2 x^2 − 3 x + 4. The graph passes the vertical line test. Thus, the equation represents a function.
- Graph y^ =^
3 x − 1 x + 2. The graph passes the vertical line test. Thus, the equation represents a function.
- (^2) x^2 + 3 y^2 = 1 Solve for y:
2 x^2 + 3 y^2 = 1 → 3 y^2 = 1 − 2 x^2 → y^2 =
1 − 2 x^2 3
y = ± 1 − 2 x^2 3
For
x = 0, y = ±
- Thus,^ 0,^
and 0,− 1 3
are on the graph. This is not a function,
since a distinct x corresponds to two different y 's.
Section 3.1 Functions
- p ( x ) =
x − 1 2 x − 1
0 → x − 1 > 0 → x > 1 Domain: (^) { x x > (^1) }
- q ( x ) = − x − 2 − x − 2 > 0 → − x > 2 → x < − 2 Domain: (^) { x x < − (^2) }
- (a) f (0) = 3 since (0,3) is on the graph. f (− 6) = −3 since (− 6,−3) is on the graph.
(b) f (6) = 0 since (6, 0) is on the graph. f (11) = 1 since (11, 1) is on the graph.
(c) f (3) is positive since f (3) ≈ 3.7.
(d) f (−4) is negative since f (−4) = −1.
(e) f ( x ) = 0 when x = − 3 , x = 6, and x = 10.
(f) f ( x ) > 0 when − 3 < x < 6 , and 10 < x ≤ 11.
(g) The domain of f is (^) { x − 6 ≤ x ≤ (^11) } or (^) [− 6, 11]
(h) The range of f is (^) { y − 3 ≤ y ≤ (^4) } or (^) [− 3 , 4]
(i) The x-intercepts are (–3, 0), (6, 0), and (11, 0).
(j) The y-intercept is (0, 3).
(k) The line y = 12 intersects the graph 3 times.
(l) The line (^) x = 5 intersects the graph 1 times.
(m) f ( x ) = 3 when x = 0 and x = 4.
(n) f ( x ) = − 2 when x = −5 and x = 8.
- (a) f (0) = 0 since (0,0) is on the graph. f (6) = 0 since (6,0) is on the graph.
(b) f (2) = −2 since (2, − 2) is on the graph. f (−2) = 1 since (−2,1) is on the graph.
Chapter 3 Functions and Their Graphs
(c) f (3) is negative since f (3) ≈ −1.
(d) f (−1) is positive since f (−1) ≈ −0.4.
(e) (^) f ( x ) = 0 when x = 0, x = 4, and x = 6.
(f) (^) f ( x ) < 0 when 2 < x < 4.
(g) The domain of f is { x − 4 ≤ x ≤ 6 } or [− 4 , 6]
(h) The range of f is { y − 2 ≤ y ≤ 3 } or [− 2 , 3]
(i) The x-intercepts are (0, 0), (4, 0), and (6, 0).
(j) The y-intercept is (0, 0).
(k) The line (^) y = − 1 intersects the graph 2 times.
(l) The line (^) x = 1 intersects the graph 1 times.
(m) (^) f ( x ) = 3 when x = 5.
(n) (^) f ( x ) = −2 when x = 2.
- Not a function since vertical lines will intersect the graph in more than one point.
- Function (a) Domain: (^) { x Real Numbers} ; Range: (^) { y y > (^0) }
(b) (0,1) (c) No symmetry to the x-axis, y-axis, or origin.
- Function (a) Domain: (^) { x − π ≤ x ≤ π} ; Range: (^) { y − 1 ≤ y ≤ (^1) }
(b) −π 2
π 2
, 0 , (0,1) (c) y-axis
- Function (a) Domain: (^) { x − π ≤ x ≤ π} ; Range: (^) { y − 1 ≤ y ≤ (^1) }
(b) (^) (− π , 0), ( π, 0), (0, 0 ) (c) origin
- Not a function since vertical lines will intersect the graph in more than one point.
- Not a function since vertical lines will intersect the graph in more than one point.
- Function (a) Domain: (^) { x x > (^0) } ; Range: (^) { y y ∈ Real Numbers}
(b) (1, 0) (c) No symmetry to the x-axis, y-axis, or origin.
Chapter 3 Functions and Their Graphs
- (^) f ( x ) = − 3 x^2 + 5 x
(a) (^) f (−1)= − 3 ( − 1 )^2 + 5 (− 1 ) ≠ − 2 (-1,2) is not on the graph of f. (b) (^) f (−2) = − 3 ( − 2 )^2 + 5 (− 2 ) (-2,-22) is on the graph of f. (c) Solve for x : − 2 = − 3 x^2 + 5 x 3 x^2 − 5 x − 2 = 0 → ( 3 x + 1 ) ( x − 2 ) = 0
→ x = −
, x = 2
(2, -2) and −
are points on the graph of f.
(d) The domain of f i s : { x x is any real number}. (e) x-intercepts: f x ( ) = 0 − 3 x^2 + 5 x = 0 → x ( − 3 x + 5 ) = 0
→ x = 0, x =
→ (0,0 ) and
(f) y-intercept: (^) f ( ) = − 0 3 0( )^2 + 5 0( ) = 0 → (0,0 )
- f^ ( x )^ =^
x + 2 x − 6 (a) f (3) =
≠ 14 (3,14) is not on the graph of f.
(b) f (4) =
= − 3 (4, –3) is the point on the graph of f.
(c) Solve for x : 2 = x + 2 x − 6 2 x − 12 = x + 2 x = 14
(14, 2) is a point on the graph of f.
(d) The domain of f is: (^) { x x ≠ (^6) }.
(e) x-intercepts: f x ( ) = 0 x + 2 x − 6
= 0 → x + 2 = 0
→ x = − 2 → −( 2,0)
(f) y-intercept: f ( ) = 0
Section 3.1 Functions
- f^ ( x )^ =^
x^2 + 2 x + 4 (a) f ( 1 ) =
1 + 4 =^
5 =^
is on the graph of^ f^.
(b) f (0) =
0 + 4 =^
4 =^
is the point on the graph of^ f^. (c) Solve for x : 1 2 =^
x^2 + 2 x + 4 x + 4 = 2 x^2 + 4 0 = 2 x^2 − x
x ( 2 x − 1 ) = 0
x = 0 or x = 1 2
and^
are points on the graph of^ f^.
(d) The domain of f is: (^) { x x ≠ − (^4) }. (e) x-intercepts: f x ( ) = 0 x^2 + 2 x + 4
= 0 → x^2 + 2 = 0 which is impossible
→ no x − intercepts
(f) y-intercept: f ( ) = 0
x^2 + 2 x + 4
- f^ ( x )^ =^
2 x^2 x^4 + 1 (a) (^) f (−1) = 2(−1)
2 (−1)^4 + 1
= 1 (–1,1) is a point on the graph of f.
(b) (^) f (2) = 2(2)
2 ( 2 )^4 + 1
is a point on the graph of^ f^. (c) Solve for x :
1 = 2 x^
2 x^4 + 1 x^4 + 1 = 2 x^2 x^4 − 2 x^2 + 1 = 0 ( x^2 −1)^2 = 0
x^2 − 1 = 0 → x = ± 1
(1,1) and (–1,1) are points on the graph of f.
(d) The domain of f is: { Real Numbers}. (e) x-intercepts: f x ( ) = 0 2 x^2 x^4 + 1
= 0 → 2 x^2 = 0 → x = 0 → (0,0 )
(f) y-intercept: f ( ) = 0
2 x^2 x^4 + 1
= 0 → (0,0 )
Section 3.1 Functions
- Solving for A:
f ( x ) = 2 x^ −^ A x − 3 and f ( 4 ) = 0
f ( 4 ) = 2(4)−^ A 4 − 3 0 =
8 − A
0 = 8 − A
A = 8
f is undefined when x = 3.
- Solve for A and B: f ( x ) = x − B x − A and^ f^ (2)^ =^0 f (1) is undefined 1 − A = 0 → A = 1
f ( 2 ) =
2 − B
0 = 2 −^ B
0 = 2 − B
B = 2
(a) III (b) IV (c) I (d) V (e) II
(a) II (b) V (c) IV (d) III (e) I
(5,2)
(6,0) (7,0)
(22,5)
(29,0) Time
Time (min)
(10,2000)
(13,4000)
(33,9000) (48,9000)
(78,19000)
- (a) Two times (b) Kevin's distance from home increased steadily at a rate of 1.5 miles per hour. (c) Kevin's distance from home did not change. (d) Kevin's distance from home decreased steadily at a rate of 10 miles per hour. (e) Kevin stayed at home for 0.2 hours. (f) Kevin's distance from home increased rapidly at the beginning and then tapered off to a very slow change in distance from home. (g) Kevin's distance from home did not change. (h) Kevin's distance from home decreased rapidly at the beginning of the interval and then tapered off as he got closer to home. (i) The furthest distance Kevin is from home is 3 miles.
- (a) Michael travels fastest between 7 and 7.4 minutes. (b) Michael's speed is zero between 4.2 and 6 minutes. (c) Between 0 and 2 minutes, Michael's speed increased from 0 to 30 miles/hour. (d) Between 4.2 and 6 minutes, Michael was stopped. (e) Between 7 and 7.4 minutes, Michael was traveling at a steady rate of 50 miles/hr. (f) Michael's speed is constant between 2 and 4 minutes, between 4.2 and 6 minutes, between 7 and 7.4 minutes, and between 7.6 and 8 minutes.
Chapter 3 Functions and Their Graphs
- (a) H (1) = 20 − 4.9(1)^2 = 20 − 4.9 = 15.1 meters H (1.1) = 20 − 4.9(1.1) 2 = 20 − 4.9(1.21) = 20 − 5.929 = 14.071 meters H (1.2) = 20 − 4.9(1.2)^2 = 20 − 4.9(1.44) = 20 − 7.056 = 12.944 meters H (1.3) = 20 − 4.9(1.3)^2 = 20 − 4.9(1.69) = 20 − 8.281 = 11.719 meters (b) H ( x ) = 15 H ( x ) = 10 H ( x ) = 5 15 = 20 − 4.9 x^2 − 5 = − 4.9 x^2 x^2 = 1. x = 1.01 seconds
10 = 20 − 4.9 x^2 − 10 = − 4.9 x^2 x^2 = 2. x = 1.43 seconds
5 = 20 − 4.9 x^2 − 15 = − 4.9 x^2 x^2 = 3. x = 1.75 seconds (c) H ( x ) = 0 0 = 20 − 4.9 x^2 − 20 = − 4.9 x^2 x^2 = 4. x = 2.02 seconds
- (a) H (1) = 20 − 13(1) 2 = 20 − 13 = 7 meters H (1.1) = 20 − 13(1.1) 2 = 20 − 13(1.21) = 20 − 15.73 = 4.27 meters H (1.2) = 20 − 13(1.2)^2 = 20 − 13(1.44) = 20 − 18.72 = 1.28 meters (b) H ( x ) = 15 H ( x ) = 10 H ( x ) = 5 15 = 20 − 13 x^2 − 5 = − 13 x^2 x^2 = 0. x = 0.62 seconds
10 = 20 − 13 x^2 − 10 = − 13 x^2 x^2 = 0. x = 0.88 seconds
5 = 20 − 13 x^2 − 15 = − 13 x^2 x^2 = 1. x = 1.07 seconds (c) H ( x ) = 0 0 = 20 − 13 x^2 − 20 = − 13 x^2 x^2 = 1. x = 1.24 seconds
- h ( x ) = − 32 x^2 1302
(a) h (100) =
−32(100)^2
- 100 = −18.93 + 100 = 81.07 feet
(b) h (300)=
−32(300)^2
- 300 = −170.41 + 300 = 129.59 feet
(c) h (500)=
−32(500)^2
- 500 = − 473.37 + 500 = 26.63 feet
(d) Solve h ( x ) =
− 32 x^2 1302
x
− 32 x 1302
x = 0 or − 32 x 1302
32 x 1302
→ x =
= 528.125 feet
Chapter 3 Functions and Their Graphs
- Let x represent the length of the rectangle.
Then x 2 represents the width of the rectangle, since the length is twice the width.
The function for the area is: A ( x ) = x ⋅ x 2
x^2 2
x^2
- Let x represent the length of one of the two equal sides.
The function for the area is: A ( x ) =
2 ⋅^ x^ ⋅^ x^ =^
2 x
2
- Let x represent the number of hours worked. The function for the gross salary is: G ( x ) = 10 x
- Let x represent the number of items sold. The function for the gross salary is: G ( x ) = 10 x + 100
- (a) The relation is not a function because 23 is paired with both 56 and 53.
- (a) The relation is not a function because 24 is paired with both 343 and 341.
(b)
18 32
65
40
(b)
18 30
355
330 (c) Using the points (20,60) and (30, 44) we get
D = −1.6 p + 92
(c) Using the points (20,335) and (28.3, 351) we get
S =1.9277 A + 296. (d) As the price of the jeans increases by $1, the demand for the jeans decreases by 1.6.
(d) As the advertising expenditure increases by $1, the sales increase by $1.9277.
(e) D ( p ) = −1.6 p + 92 (e) (^) S A ( ) =1.9277 A + 296. (f) (^) Domain: (^) { p p > (^0) } (f) (^) Domain: (^) { A A ≥ (^0) } (g) D (28) = −1.6(28) + 92 = 47.2 ≈ 47 Demand is about 47 pairs.
(g) S (25) =1.9277(25) + 296. ≈ 344. Sales are about $344,638. (h) D^ = −1.3355 p^ +^ 86.1974^ (h) S^ =^ 2.0667^ A^ +^ 292.
Section 3.1 Functions
- (a) The relation is a function. Each time value is paired with exactly one distance value.
- (a) The relation is a function. Each HS GPA value is paired with exactly one College GPA value. (b)
0 10
350
0
(b)
2.5^4
4
(c) Using the points (0,0) and (8, 300) we get
s = 37.5 t
(c) Using the points (2.73,2.43) and (3.10, 2.93) we get
G = 1.3514 x −1. (d) As the time increases by 1 hour, the distances increases by 37. miles.
(d) As the high school GPA increases by 1 , the college GPA increases by 1.3514.
(e) s ( t )^ =^ 37.5 t^ (e) G x ( ) =1.3514 x −1. (f) (^) Domain: (^) { t t ≥ (^0) } (f) (^) Domain: (^) { x x ≥ (^0) } (g) s (11)^ =^ 37.5(11) = 412.5 miles
(g) G (3.23)^ =^ 1.3514(3.23)^ −1. = 3.1058 ≈ 3.
The college GPA is about 3.12. (h) s = 37.7833 t − 19.1333 (h) (^) G = 0.9639 x + 0.
- (a) h ( x ) = 2 x h ( a + b ) = 2( a + b ) = 2 a + 2 b = h ( a ) + h ( b ) h ( x ) = 2 x has the property. (b) g ( x ) = x^2 g ( a + b ) = ( a + b )^2 = a^2 + 2 ab + b^2 ≠ a^2 + b^2 = h ( a ) + h ( b ) g ( x ) = x^2 does not have the property. (c) F ( x ) = 5 x − 2 F ( a + b ) = 5( a + b ) − 2 = 5 a + 5 b − 2 ≠ 5 a − 2 + 5 b − 2 = h ( a ) + h ( b ) F ( x ) = 5 x − 2 does not have the property. (d) G ( x ) =
x G ( a + b ) =
a + b
a
b = h ( a ) + h ( b )
G ( x ) =
x does not have the property.
- All points of the form (5, y) and of the form (x, 0) cannot be on the graph of the function.
- No, x = − 1 is not in the domain of g , but it is in the domain of f.
- Answers will vary
- A function may have any number of x-intercepts; it can have only one y-intercept.
- Yes. One example is ,where the domain of is {1}.
- A function cannot be symmetric to the x-axis, because it would then fail the vertical line test.
G ( x ) = 2 G
