Math 121.
Written Homework 7, Fall 2016
Due. Tuesday, December 6
Instructions. Do each of the following nine problems, you must use the appropriate methods from
algebra and show your work to receive maximal credit. This assignment can only help your grade! It
will be used to replace your lowest written assignment.
1. A function f that is one-to-one on its domain is graphed below.
8 y
6
4
2
x
−8 −6 −4 −2
−2
2
4
6
8
−4
−6
−8
For your reference, the points plotted on the graph of f are
(3, −5), (2, −4), (−1, −3), (−6, −2)
(a) Sketch the graph of f −1 on the same graph.
(b) Find (i) f −1 (−2) and (ii) f −1 (−5).
(c) Find (i) the domain of f −1 and (ii) the range of f −1 .
Solution: (a) The graph of f −1 is the reflection of the graph of f over the line y = x. This is
given below.
8 y
6
4
2
x
−8 −6 −4 −2
−2
2
4
6
8
−4
−6
−8
(b) (i) f −1 (−2) = −6 because f (−6) = −2 and (ii) f −1 (−5) = 3 because f (3) = −5.
(c) (i) The domain of f −1 is [−5, ∞) because this is the range of f .
(ii) The range of f −1 is (−∞, 3] because this is the domain of f .
2. (a) Find the inverse function of f (x) =
7x + 2
, and find its domain.
8x − 2
(b) Find the range of f .
7x + 2
= 2, if possible. Note. This is just asking you to solve f (x) = 2 for x.
(c) Solve the equation
8x − 2
7x + 2
7
(d) Solve the equation
= , if possible.
8x − 2
8
7x + 2
, and then switch x and y to have
8x − 2
Solution: (a) We first write y =
x=
7y + 2
8y − 2
Then we will solve the preceding expression for y to obtain the inverse function of f .
x(8y − 2) = 7y + 2 ⇒ 8xy − 2x = 7y + 2 ⇒ 8xy − 7y = 2x + 2
Thus
2x + 2
y=
8x − 7
7
7
2x + 2
−1
−1
for x 6= , and the domain of f = x : x 6=
.
Thus f (x) =
8x − 7
8
8
7
−1
(b) The range of f is the domain of f , i.e., the range of f is x : x 6=
.
8
y(8x − 7) = 2x + 2
and so
(c) It is possible to solve this, because 2 is in the domain of f −1 (x), and f (x) = 2 is equivalent
to x = f −1 (2), therefore, using (a)
x=
(d) This is not possible to solve, because
2(2) + 2
6
2
= = .
8(2) − 7
9
3
7
8
is not in the range of f (see part (b)).
3. Consider the function f (x) = 2x
(a) Complete the following table of values for f .
−3
x
−2
−1
0
1
2
3
y
(b) Sketch a graph of f , and on the same coordinate axes sketch y = x and the graph of f −1 (x)
(c) Using your answer from (b), sketch g(x) = 2x+3 − 6.
Solution: (a) The completed table is as follows.
x
−3
−2
−1
0
1
2
3
y
1/8
1/4
1/2
1
2
4
8
In (b), recall that the graph of the inverse will be the graph of f reflected over the line y = x. In
Page 2
(c), the graph of y = 2x is shifted 3 units to the left and 6 units down to obtain the graph of g.
(b)
(c)
8 y
8 y
6
6
f −1
4
4
2
f
2
x
−8 −6 −4 −2
−2
2
4
6
x
−8 −6 −4 −2
−2
8
−4
g
2
4
6
8
−4
−6
−6
−8
−8
4. The graph of an exponential function y = bx is given on the graph below.
8 y
6
4
2
x
−8 −6 −4 −2
−2
2
4
6
8
−4
−6
−8
(a) Use the graph to estimate b.
(b) Using the graph from (a), graph y = b|x| .
(c) Using your graph from (b), graph y = b|x+3| − 4.
Solution: (a) b =
1
2
because the graph goes through the point (−1, 2).
(b) This graph results in reflecting the part of the original graph for x ≥ 0 over the y-axis.
(c) The requested graph results from shifting the original graph 3 units to the left and 4 units
down. The graph is given below
Page 3
8 y
6
4
2
−8 −6 −4 −2
−2
(c)
−4
(b) x
2
4
6
8
−6
−8
5. The population of a small city is currently 52000 and is growing at 5 percent per year. Thus the
population is given by
P (t) = 52000(1.05)t
where t is time measured in years from the present.
(a) What will the population of the city be in one year?
(b) According to this model, what will the population of the city be in 13 years from now? Express
answer to the nearest whole number.
(c) Suppose Charles has an investment account that is growing a a rate of 5 percent per year, and he
currently has 52000 dollars in the account. How much money will be in the account 13 years from
now? Express answer to the nearest dollar.
Solution: (a) P (1) = 52000(1.05) = 54600 which represents an increase of 5 percent over the
current populaton.
(b) Assuming the current growth rate continues, the population will be
P (t) = 52000(1.05)13 = 98054.
(c) This is mathematically the same question as (b), except expressed in dollars, not population,
so the account will have 52000(1.05)13 = 98054 dollars in it after 13 years.
Page 4
1 x
5
and g(x) = log 1 x
5
x
(a) Complete the following table of values for f (x) = 15 .
6. Consider the functions f (x) =
−3
x
−2
−1
0
1
2
3
1 x
5
(b) Complete the following table of values for g(x) = log 1 x by filling in the x-values for the given
5
y-values.
x
log 1 x
−3
−2
−1
0
1
2
3
5
What do you notice about the table of (a) and (b)?
x
(c) Sketch the graphs of f (x) = 15 and g(x) = log 1 x on the same coordinate axes.
5
(d) Using your answer from (b), sketch h(x) = log 1 (x − 1) + 2.
5
Solution: (a) The completed table is as follows.
x
−3
−2
−1
0
1
2
1 x
5
125
25
5
1
1/5
1/25
3
1/125
(b) Because g(x) is the inverse function of f (x), it switches the x and y values from f , so its table
just switches the rows of the table in (a).
x
125
25
5
1
1/5
1/25
1/125
log 1 x
−3
−2
−1
0
1
2
3
5
In (c), recall that the graph of the inverse g will be the graph of f reflected over the line y = x
which is the dashed line in the graph below. In (d), the graph of g(x) is shifted 1 units to the
right and 2 units up to obtain the graph of h.
(c)
(d)
f8 y
8 y
6
6
4
4
2
2
−8 −6 −4 −2
−2
2
4
6
g
h
x
8
−8 −6 −4 −2
−2
−4
−4
−6
−6
−8
−8
Page 5
2
4
6
x
8
7. The graph of a function f (x) = logb x is given in both graphs below.
(a)
(b)
8 y
8 y
6
6
4
4
f
2
f
2
x
−8 −6 −4 −2
−2
2
4
6
x
−8 −6 −4 −2
−2
8
−4
−4
−6
−6
−8
−8
2
4
6
8
(a) Sketch the graph of g(x) = logb |x| − 3
(b) Sketch the graph of h(x) = −2 logb x
(c) Use the given graph to find the base b.
Solution: (c) Notice that base is b = 5 because the graph goes through the point (5, 1).
The requested graphs are as follows. Notice for (a) the graph will be made symmetric about the
y-axis, and shifted down 3 units. For (b), the graph will be reflected over the x-axis and then
vertically stretched by a factor of 2.
(a)
(b)
8 y
8 y
6
6
4
4
2
2
x
−8 −6 −4 −2
−2
2
4
6g 8
−8 −6 −4 −2
−2
−4
−4
−6
−6
−8
−8
Page 6
h
2
x
4
6
8
36y 9
in terms of log6 x, log6 y, log6 z and
8. Use properties of logarithms to fully expand log6 √
x5 z 3
number(s); simplify your answer (assume x > 0, y > 0, and z > 0).
Solution:
log6
36y 9
√
x5 z 3
= log6 (36y 9 ) − log6
√
x5 z 3
1
log6 (x5 z 3 )
2
1
= log6 (62 ) + 9 log6 (y) − [log6 (x5 ) + log6 (z 3 )]
2
3
5
= 2 + 9 log6 y − log6 x − log6 z.
2
2
= log6 (36) + log6 (y 9 ) −
9. (Richter Scale) The magnitude of an earthquake of intensity I on the Richter scale is
I
M = log
I0
where I0 is the intensity of a zero-level earthquake.
(a) Find the magnitude to the nearest 0.1 of an earthquake that has an intensity of I = 125892541I0 .
I
in exponential form, and then solve for I.
(b) Write the formula M = log
I0
(c) Use the formula in (b) to find the intensity of an earthquake that measures 4.1 on the Richter
scale.
(d) How many times more intense is an earthquake with a magnitude of 8.5 than an earthquake with
a magnitude of 4.5? Express answer to nearest whole number. (To do this, find the intensity of each
earthquake using your formula from (b), and then divide to find the ratio of the intensities).
Solution: (a) The magnitude is M = log
(b) The exponential form is 10M =
given its magnitude.
125892541I0
I0
= log(125892541) ≈ 8.1.
I
, and so I = 10M I0 provides the intensity of an earthquake
I0
(c) From the formula in (b), the intensity is I = 104.1 I0 ≈ 12589I0 .
M1
108.5 I0
=
= 10(8.5−4.5) = 104.0 ≈ 10000. Therefore, a magnitude 8.5 earthquake is
M2
104.5 I0
approximately 10000-times as intense as a magnitude 4.5 earthquake.
(d)
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