Section 6.1 answers
1a)
1c)
3a)
3c)
5a)
5c)
(𝑓 ∘ 𝑔)(𝑥) = 6x+24
(𝑔 ∘ 𝑓)(𝑥) = 6x-2
(𝑓 ∘ 𝑔)(𝑥) = 3(3x2 – 8x + 7)
(𝑔 ∘ 𝑓)(𝑥) = 3x2 + 11
(𝑓 ∘ 𝑔)(𝑥) = x2 + 2x – 5
(𝑔 ∘ 𝑓)(𝑥) = x2 – 6x + 7
1b)
1d)
3b)
3d)
5b)
5d)
domain (−∞, ∞)
domain (−∞, ∞)
domain (−∞, ∞)
domain (−∞, ∞)
domain (−∞, ∞)
domain (−∞, ∞)
2(𝑥−7)
7a) (𝑓 ∘ 𝑔)(𝑥) = 4𝑥−25
7b) domain of (𝑓 ∘ 𝑔)(𝑥) is all real numbers except 7 and 25/4.
3(𝑥+4)
7c) (𝑔 ∘ 𝑓)(𝑥) =−7𝑥+26
7d) domain of (𝑔 ∘ 𝑓)(𝑥) is all real numbers except -4 and 26/7
𝑥
9a) (𝑓 ∘ 𝑔)(𝑥) = −3𝑥+1
9b) : domain of (𝑓 ∘ 𝑔)(𝑥) is all real numbers except 0 and 1/3
9c) (𝑔 ∘ 𝑓)(𝑥) =x-3
9d) domain of (𝑔 ∘ 𝑓)(𝑥) is all real numbers except 3
11a) show (𝑓 ∘ 𝑔)(𝑥) = x
11b) show (𝑔 ∘ 𝑓)(𝑥)
(𝑓 ∘ 𝑔)(𝑥) =f(g(x))
(𝑓 ∘ 𝑔)(𝑥) = 7(𝑔(𝑥)) + 1
(𝑔 ∘ 𝑓)(𝑥)
(𝑓 ∘ 𝑔)(𝑥) =
=x–1+1
=x
𝑥−1
7( 7 )+
(𝑔 ∘ 𝑓)(𝑥)
1
= g(f(x))
𝑔(𝑥)−1
7
(𝑔 ∘ 𝑓)(𝑥) =
13a) show (𝑓 ∘ 𝑔)(𝑥) = x
(𝑓 ∘ 𝑔)(𝑥) =
𝑔(𝑥)−5
2
(𝑓 ∘ 𝑔)(𝑥) =
2𝑥+5−5 2𝑥
= 2
2
=x
=x
(7𝑥+1)−1
7
=
7𝑥+1−1 7𝑥
=
7
𝑥
13b) 𝑠ℎ𝑜𝑤 (𝑔 ∘ 𝑓)(𝑥)
=x
(𝑔 ∘ 𝑓)(𝑥)
= 2(𝑓(𝑥)) + 5
(𝑔 ∘ 𝑓)(𝑥)
= 2(
𝑥−5
)+
2
=x
5 = x-5+5 = x
Section 6.2 answers
1) not one to one
3) one to one
5) one to one
9) not one to one
11) one to one
13) not one to one
-1
15) is one to one, h = { (3,0) (1,5) (11,7) (-3,9)
17) is one to one m-1 = { (2,0) (3,2) (5,3) }
19) points labeled (-3,0) (3,1) and (9,2)
23)
The new graph should have points
(1/8, ½) (-1/8, -1/2) (8,2) (-8,-2)
7) not one to one
21) points labeled (8,-1) (2,0) (-4,1)
25) . The new graph should have points (0,0)
(2,4) (1,1) (3,9)
Section 6.2 continued
27a) 𝑓 −1 (𝑥) =
𝑥+4
2
27b)
(𝑓 ∘ 𝑓 −1 )(𝑥) = 𝑓(𝑓 −1 (𝑥))
(𝑓 ∘ 𝑓 −1 )(𝑥) = 2(𝑓 −1 (𝑥)) − 4
(𝑓 ∘ 𝑓 −1 )(𝑥)
𝑥+4
=2( 2 )−
4 = x+ 4 – 4 = x
27c) inverse drawn with dashed line
(𝑓 −1 ∘ 𝑓)(𝑥)
= 𝑓 −1 (𝑓(𝑥))
𝑓(𝑥)+4
2
2𝑥−4+4
= 2
(𝑓 −1 ∘ 𝑓)(𝑥) =
(𝑓 −1 ∘ 𝑓)(𝑥)
=
2𝑥
2
=x
Section 6.2 continued
29a) f-1(x) = 3x+2
29b)
(𝑓 ∘ 𝑓 −1 )(𝑥) =𝑓(𝑓 −1 (𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥)
𝑓−1 (𝑥)−2
(𝑓 ∘ 𝑓 −1 )(𝑥) =
(𝑓 ∘ 𝑓 −1 )(𝑥) =
3
3𝑥+2−2
3
= 𝑓 −1 (𝑓(𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥) = 3(𝑓(𝑥)) + 2
=
3𝑥
=
3
x
29c) inverse drawn with dashed line
(𝑓 −1 ∘ 𝑓)(𝑥)
=3
𝑥−2
3
+ 2 = x-2+2 = x
Section 6.2 continued
31a) f −1 (x) =
2
x
31b)
(𝑓 ∘ 𝑓 −1 )(𝑥) =𝑓(𝑓 −1 (𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥)
2
𝑓−1 (𝑥)
2
(𝑓 ∘ 𝑓 −1 )(𝑥) =
(𝑓 ∘ 𝑓
−1 )(𝑥)
=
2
2
𝑥
= 2∗
= 𝑓 −1 (𝑓(𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥) =
𝑓(𝑥)
𝑥
2
(𝑓 −1 ∘ 𝑓)(𝑥) =
=x
31c) only one graph shown as the function is its own inverse.
2
2
𝑥
𝑥
= 2 ∗ 2 =x
Section 6.2 continued
33a) f-1(x) = x3 + 4
33b)
(𝑓 ∘ 𝑓 −1 )(𝑥) =𝑓(𝑓 −1 (𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥)
= 𝑓 −1 (𝑓(𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥) = (𝑓(𝑥))3 + 4
3
(𝑓 ∘ 𝑓 −1 )(𝑥) = √𝑓 −1 (𝑥) − 4
3
(𝑓 −1 ∘ 𝑓)(𝑥) = 3√𝑥 − 4 + 4 = x-4 + 4 = x
3
3
(𝑓 ∘ 𝑓 −1 )(𝑥) = √𝑥 3 + 4 − 4 = √𝑥 3 = x
33c)
Section 6.2 continued
3
35a) 𝑓 −1 (𝑥) = √𝑥 − 2
35b)
(𝑓 ∘ 𝑓 −1 )(𝑥) =𝑓(𝑓 −1 (𝑥))
(𝑓 ∘ 𝑓 −1 )(𝑥) = (𝑓 −1 (𝑥))3 + 2
(𝑓 −1 ∘ 𝑓)(𝑥)
= 𝑓 −1 (𝑓(𝑥))
(𝑓 −1 ∘ 𝑓)(𝑥) = 3√𝑓(𝑥) − 2
3
(𝑓 ∘ 𝑓
35c)
−1 )(𝑥)
3
3
= √𝑥 − 2 + 2 = x-2 + 2 = x
3
(𝑓 −1 ∘ 𝑓)(𝑥) = √𝑥 3 + 2 − 2 = √𝑥 3 = x
Section 6.3 answers
1) f(x) = 2x
a) make a table of values and sketch a graph
x
2
1
0
-1
-2
f(x)
22 = 4
21 = 2
20 = 1
2-1 = ½
2-2 = ¼
(2,4)
(1,2)
(0,1)
(-1, ½)
(-2, ¼)
1b) domain (−∞, ∞)
1c) 𝑟𝑎𝑛𝑔𝑒 (0, ∞)
1d) y = 0
3) f(x) = 2x+3
a) make a table of values and sketch a graph
x
-5
-4
-3
-2
-1
f(x)
2-5+3 = 2-2 = =1/22 = 1/4
2-4+3 = 2-1 = ½
(-4, ½)
2-3+3 = 20 = 1 (-3, 1)
2-2+3 = 2
(-2, 2)
2-1+3 = 22 = 4 (-1, 4)
3b) domain (−∞, ∞)
(-5, 1/4)
3c) 𝑟𝑎𝑛𝑔𝑒 (0, ∞)
3d) y = 0
Section 6.3 answers
5a) f(x – 3) = ex – 3
5b) The graph is the same, but shifted 3 units to the right.
5c) Graph of f(x – 3) drawn in blue.
5d) domain (−∞, ∞)
5e) 𝑟𝑎𝑛𝑔𝑒 (0, ∞)
5f) y = 0
e) The graph is the same, but shifted 5 units to the right.
7a) f(x + 2) = ex + 2
7b) The graph has the same shape, but is shifted 2 units to the left.
7c) Graph of f(x + 2) is drawn in blue.
7d) domain (−∞, ∞)
7e) 𝑟𝑎𝑛𝑔𝑒 (0, ∞)
7e) The graph is the same, but shifted 4 units to the left.
7f) y = 0
Section 6.3 answers
9a) f(x) + 2 = ex + 2
9b) Shift up 2 units.
9c) f(x) + 2 is drawn in blue, and the horizontal asymptote is drawn in purple.
9d) domain (−∞, ∞)
9e) 𝑟𝑎𝑛𝑔𝑒 (2, ∞)
9f) y = 2
Section 6.3 answers
11a) f(x) – 3 = ex – 3
11b) shifted down 3 units.
11c) graph of f(x) – 3 drawn in blue, horizontal asymptote drawn in purple.
11d) domain (−∞, ∞) 11c) 𝑟𝑎𝑛𝑔𝑒 (−3, ∞) 11d) y = -3
Section 6.3 answers
13a) f(-x) = e-x
13b) reflect over y-axis
13c) graph of f(-x) drawn in blue.
13d) domain (−∞, ∞)
13e) 𝑟𝑎𝑛𝑔𝑒 (0, ∞)
13f) y = 0
15a) 2f(x) = 2ex
15b) graph is stretched
15c) graph of 2f(x) drawn in blue.
15d) domain (−∞, ∞)
Multiply each y in the original table of f(x) by 2 to get the yvalues for the points in 2f(x)
x
f(x)
2f(x) y computation
Point on graph of
2f(x)
1
-2
(-2, 0.28)
= 0.14 2*0.14 = 0.28
-1
𝑒2
1
=0.37
𝑒
2*0.37 = 0.74
(-1, 0.74)
0
1
2
1
e = 2.72
e2 = 7.39
2*1 = 2
2*2.72 = 5.44
2*7.39 = 14.78
(0, 2)
(1, 5.44)
(2, 14.78)
15e) 𝑟𝑎𝑛𝑔𝑒 (0, ∞)
15f) y = 0
Section 6.3 continued
17a) g(x + 1) = 2x + 1
17b) shifts left 1
19a) g(x – 1) = 2x – 1
19b) shifts right 1
21a) g(x) + 1 = 2x + 1
21b) shifts up 1
23a) g(x) – 2 = 2x – 2
23b) shifts down 2
25a) –g(x) = -2x
25b) reflects over x-axis
27a) g(x + 1) – 4 = 2x + 1 – 4
27b) shifts left 1 and down 4
29a) g(x – 2) + 3 = 2x – 2 + 3
29b) shifts right 2 and up 3
31a) –g(-x) + 2 = -2-x + 2
31b) reflects over x-axis, reflects over y-axis, shifts up 2
33) x = 1
35) x = -3
37) x = 2
39) x = -5/2
41) x = 2
43) x = -10
45) x = -1
47) x = -2
49) x = 6
Section 6.4 Answers
1
3
1) log3(9) = 2
3) log3(81) = 4
5) 𝑙𝑜𝑔3 = −1
7) ln(x) = y
9) ln(20.09) = 3
11)ln(2.72) = 1
13) 34 = 81
15) 26 = 64
17) 61 = 6
19) 103 = x
21) e1 = x
23) ew = 2x
25) e2 = e2
27) 1
29) 1
31) 0
33) 0
35) 3
37) 1
39) 0
41) 2
43) 7
45) 3
47) 6
49) 5
51) 1
53) 3
55) -2
57) .7782
59) -.4771
61) -2.0969
63) 1.9459
65) 2.7726
67a)
x
32
31
30
3-1
3-2
y
2
1
0
-1
-2
point
(9,2)
(3,1)
(1,0)
(1/3, -1)
(1/9, -2)
b) State the domain of each function.
67b) domain: (0, ∞)
69a)
Create a table of values, I will put the numbers 2,1,0,-1,-2 in the y column and solve for x.
x
e2
e1
e0
e-1
e-2
y
2
1
0
-1
-2
point
(7.39, 2)
(2.72, 1)
(1, 0)
(.37, -1)
(.14, -2)
69b) domain (0, ∞)
71a)
x
y
2
point
(1/9, 2)
1
(1/3, 1)
0
(1, 0)
(3)
-1
(3, -1)
( )
-2
(9,-2)
1 2
(3)
1 1
( )
3
1 0
(3)
1 −1
1 −2
3
71b) domain (0, ∞)
73a)
73b)
73c)
73d)
f(x + 1) = log2(x + 1)
x > -1 or (−1, ∞)
shifts left 1
Graph of f(x + 1) drawn in blue
75a) f(x – 2) = log2(x – 2)
75b) x > 2 or (2, ∞)
75c) Shifts right 2
75d)
77a) f(x) + 2 = log2(x) + 2
77b) x > 0 or (0, ∞)
77c) shifted up 2 units
77d) Just shift each point in the graph of f(x) two units to the up. I showed the x>0 domain as a vertical
asymptote drawn in purple. The graph will not exist to the left of this vertical line x = 0.
Graph of f(x) + 2 drawn in blue
79a) f(x) – 2 = log2(x) – 2
79b) x > 0 or (0, ∞)
79c) Shifted down 2 units
79d) Just shift each point in the graph of f(x) two units to the down. I showed the x>0 domain as a
vertical asymptote drawn in purple. The graph will not exist to the left of this vertical line x = 0.
Graph of f(x) - 2 drawn in blue
81a) f(-x) = log2(-x)
81b) any of these three answers are correct, you only need to give one answer.
0 > x or x < 0 or (−∞, 0)
81c) reflects over the y-axis
81d) Just reflect each point over the y-axis. The graph will now only exist to the left of the y-axis. The
vertical asymptote will still be at x = 0 (or the y-axis). It’s now the right edge of the graph as opposed to
the left edge of the graph.
83a) 3f(x) = 3log2(x)
83b) x > 0 or (0, ∞)
83c) stretches the graph
83d) This is a non-rigid transformation. I need to make a table of values to sketch an accurate graph.
We can use the x’s from the given table. We create y’s by multiplying each y-value by 3.
3f(x) is drawn in blue. The vertical asymptote is
drawn in purple.
Here are the points that are marked in the
original graph
x
f(x)
.25
-2
.5
-1
1
0
2
1
4
2
The table for 3f(x) will have the same x-values,
but the y’s will be multiplied by 3.
Here is the table for 3f(x)
x
3f(x)
.25
-6
.5
-3
1
0
2
3
4
6
Section 6.5 answers
1) 4
3) 6
5) 5
7) 5
9) 1
11) 15
13) 2log3x + 3log3y
15) 2log5x + 6log5y + log5z
17) log2x + 3log2y – 2log2z
19) log2x + log2 y - 2log2w - 5log2z
1
3
23) 2𝑙𝑜𝑔2 𝑥 + 𝑙𝑜𝑔2 𝑦
29) 𝑙𝑜𝑔2
𝑥 5𝑦3
𝑧
21) 3log4x + 4log4y
25) log2 (x3y4)
27) log3 (x2y4z)
𝑥4
𝑦𝑧
31) 𝑙𝑜𝑔 𝑦2 𝑧3
33) 𝑙𝑜𝑔3 𝑥2
37) 1.58
39) 1.46
41) -.31
43) 1.80
45) -.95
47) 17
49) 40
51) 17
53) 65
55) 10
57) -26
35)
𝑙𝑛
𝑥𝑦 3
𝑧2
Section 6.6 answers
1) x = 4
3) x = 4
5) x = -4
7) x = -2
9) x = 1/5
11) x = 1/2
13) x= Log3 6 = 1.63
15) x = ln(12) = 2.48
17) x = log(4) = .60
19) x =
ln(4)
2
= .69
log(3)
log(6)
21) 𝑥 = log(3)−log(5) = −2.15
23) 𝑥 = log(6)−log(9) = −4.42
25) x = 9
27) x = e1 = 2.72
29) x = 7
31) x = 4
33) x = 9
35) x = 9
37) x = 16
39) x = 3/2
41) x = 3/2
43) x = 2
45) x = 8
47) x = 10
49) Answer: x = 3
(no need to mention the x =-9 as it doesn’t check)
It would be okay to write: Answer: x = 3, x = -9 is extraneous)
51) Answer: x = 3
(no need to mention the x =-9 as it doesn’t check)
It would be okay to write: Answer: x = 3, x = -9 is extraneous)
Section 6.7 answers:
1) The investment will be worth $1,218.99 in 5 years
3) The investment will be worth $17,595.65 in 8 years.
5) The investment will be worth $6,749.29 in 5 years.
7) The investment will be worth $17,958.26 in 6 years.
9) 18.3 years
11) 57.6 years
13) The house will be cost $231,854.81 in 5 years.
15) The house will be cost $106,120.80 in 3 years.
17) Tuition will be $97.33 per credit hour in 5 years
Section 6.8 answers
1a) 19 people
1b) 22%
1c) 24 days
1d) it will take 1.96 years for the number of text messages to triple
3a) the initial population is 200 insects
3b) The growth rate is 4%
3c) there will be 265 insects after 7 days.
3d) 67.70 days
3e) 17.33 years
5a) decay rate 6.3% per year
5b) 151.07 years
5c) 25.55 years
5d) 11 years
7) 3.97 years
9) 29.26 years
11) 7.4 years
13) 9.97 years
Section 6.9 answers
1a)
1b) Exponential Regression
1c) P(t) = 172.544e-.012t
1d) 33.87 minutes
3c) P(t) = 9.696e.767t
3d) 6.95 days
3a)
3b) exponential regression
Section 6.9 continued
5a)
5b) exponential regression
5c) P(t) = 49.616e-.130t
5d) 12.30 weeks
7a)
7b) logarithmic regression
7d) 9.36 years
7c) y = 5.946 + 6.286ln(t) or P(t) = 5.946 + 6.286ln(t)
Chapter 6 review answers
1a) (𝑓 ∘ 𝑔)(𝑥) = 2(8x2 + 1)
1b) domain (−∞, ∞)
2a) (𝑓 ∘ 𝑔)(𝑥) = 3x – 12
2b) domain all real numbers except 4
3) f is one to one (since all y’s are different)
f-1 = { (1,0) (4,1) (0,2) (5,3)}
4) g is not one to one as there are two points
with the same y value.
No need to find g inverse as it would not be a
function.
5) Simply switch the three points that given. Plot the points (0,-3) (2,1) and (3,6) and connect them with
the same shape. I also drew in the line y = x as the graphs are supposed to be mirrored images over that
line.
6a) 𝑓−1 (𝑥) = √𝑥 + 5
3
6b)
(𝒇 ∘ 𝒇−𝟏 )(𝒙) = 𝒇(𝒇−𝟏 (𝒙))
𝟑
(𝒇 ∘ 𝒇−𝟏 )(𝒙) = (𝒇−𝟏 (𝒙)) − 𝟓
𝟑
𝟑
(𝒇 ∘ 𝒇−𝟏 )(𝒙) = (√𝒙 + 𝟓) − 𝟓 = 𝒙 + 𝟓 − 𝟓 = 𝒙
(𝒇−𝟏 ∘ 𝒇)(𝒙)
𝟑
𝟑
= √𝒙𝟑 − 𝟓 + 𝟓 = √𝒙𝟑 = 𝒙
Chapter 6 review answers:
7a) g(x+5) = 2x+5 shift left 5 units
7b) g(x-4) = 2x-4 shift right 4 units
7c) g(x) + 1 = 2x + 1 shift up 1 unit
7d) g(x) -3 = 2x - 3 shift down 3 units
7e) -g(x) = -2x: reflected over x-axis
7f) g(-x) = 2-x reflected over y-axis
7g) g(x+3)+5 = 2x+3 +5 shift left 3 units up 5 units
7h) g(x-3)-2 = 2x-3 - 2 shift right 3 units down 2 units
7i) -g(x+1) = -2x+1 reflect over x-axis, shift left 1 unit
7j) -g(-x) = -2-x reflect over x-axis and reflect over y-axis
8) x = 6
9) x = 3
10) x = 5
11) x = 3
12a)
x
22 = 4
21 = 2
20 = 1
2-1 = 1/2
2-2 = 1/4
12b) domain (0, ∞)
y
2
1
0
-1
-2
point
(4,2)
(2,1)
(1,0)
(1/2, -1)
(1/4, -2)
Chapter 6 review answers:
13a) f(x+1) = log2(x+1)
13b) x > -1 or (−1, ∞)
13c) Shifts left 1
13d) graph of log2(x+1) drawn in blue.
14a) f(x) – 1 = log2(x) – 1
14b) x > 0 or (0, ∞)
14c) Shifts down 1
14d) Shift every point down 1 unit. The graph of f(x) – 1 is drawn in blue. The y-axis is the vertical
asymptote. It has an equation x = 0 and I graphed it in purple.
15a) f(x – 3) + 2 = log2(x-3) + 2
15b) x > 3 or (3, ∞)
15c) shifts right 3 and up 2
15d) Shift each point right 3 and up 2. The vertical asymptote moves over 3 to the right. There will be a
vertical asymptote at x = 3. It is drawn in purple. The graph of f(x – 3) + 2 is drawn in blue.
16a) -2f(x) = -2log2(x)
16b) x > 0 or (0, ∞)
16c) reflected over the x-axis and stretched.16d) I had to make a table of values because of the
stretching. I used the x’s from the given graph and multiplied each y by -2 to get the y’s for the desired
graph. The graph of -2f(x) is drawn in blue. There is a vertical asymptote at x = 0 which is drawn in
purple.
Here are the points in the given graph
x
f(x)
.25
-2
.5
-1
1
0
2
1
4
2
Points in graph of -2f(x)
just multiply each y by -2
x
f(x)
.25
4
.5
2
1
0
2
-2
4
-4
Chapter 6 review answers
𝑥3
17) 𝑙𝑜𝑔2 𝑦4
18) 𝑙𝑛
21) 2ln(x) + ln(y) – ln(z)
𝑥5𝑧
𝑦2
𝑥2
19) 𝑙𝑜𝑔3 𝑦4 𝑧
20) log3y - 2log3x – log3z
22) 3log4x + 2log4y + 5log4z
23) x = 2.26
24) x = -7.03
25) x = 81
26) x = 7.39
27) x = 31
28) x = 10
29) x = 2
30) x = 2
35) x = 2
or x = 2, x = -10 is extraneous
36) x = 3
or x =3. x = -9 is extraneous
31) 36.6 years
32) 94.67 years