Find the inverse of f(x) = 5x + 15.
y = 5x + 15
inverse:
x = 5y + 15
-5y = -x + 15
y = (1/5)x - 3
f-1(x) = (1/5)x - 3
y
The inverse of a function is a
reflection over the line y = x.
2
2
-2
-2
x
Each point in the inverse is a
transposition of the point in the
original function. (The domain and
the range are interchanged).
Given f(x) = 2x and g(x) = log2 x.
a. Plot the graph of f in the domain D = {x | -3 ≤ x ≤ 3}.
b. Plot the graph of g in the domain D = {x | 0.125 ≤ x ≤ 8}.
c. Describe the relationship between the two graphs.
y
x
f(x)
x
g(x)
-3
0.125
0.125 -3
-2
0.25
0.25
-2
-1
0.5
0.5
-1
0
1
1
0
1
2
2
1
2
x
2
4
4
2
3
8
8
3
-2
2
-2
Find the inverse of f(x) = x2. Then tell whether or not the inverse is a
function.
y = x2
x = y2
y
x = y2
x= y
2
y=± x
x
-2
2
-2
Algebraic method for determining inverses:
Use rules of composite functions.
Check does f(g(x)) = g(f(x)) = x.
Are f(x) = 5x + 15 and g(x) = (1/5)x - 3 inverses of each other?
æ1
ö
f (g( x)) = f ç x - 3÷ =
è5
ø
æ1
ö
5ç x - 3÷ + 15
è5
ø
y
= ( x - 15) + 15 = x
g( f ( x)) = g(5x + 15) = 1 (5x + 15) - 3
5
= ( x + 3) - 3 = x
2
2
-2
-2
x
If possible, classify the type of function in each table below.
x
y
x
y
x
y
1
12
5
20
2
10
4
48
10
40
9
73
7
192
15
60
16
234
10
768
20
80
23
493
Exponential, as
you add 3 to x, you
multiply y by 4.
10
73
Linear, as you add
5 to x, you add 20
to y.
234
493
73
234
493
-10
-73
- 234
63
161
259
161
259
-63
- 161
98
98
Quadratic (this
uses finite
differences).
After 2 finite
differences, the results
are the same,
therefore it is
quadratic. (3 finite
differences would be
cubic.)
Assume the population of the US is increasing exponentially with time. The
1970 census showed that the population was about 203 million. The 1980
census showed that the population had grown to about 226 million.
1. What was the population in 1990, and 2000?
Let x = years since 1970, then
Let y = population of the US in millions
(Use the add-multiply property of exponential equations.)
+10
x
y
0
203
10
226
+10
(1.113300493)
20
+10
(1.113300493)
30
251.6
280.1
(1.113300493)
The population in 1990 was about 251.6 million, and the
population in 2000 was about 280.1 million.
Assume the population of the US is increasing exponentially with time. The
1970 census showed that the population was about 203 million. The 1980
census showed that the population had grown to about 226 million.
2. Find a particular equation expressing population in terms of the
number of years that have elapsed since 1970.
y = abx
Remember, we have two given points: (0, 203) and (10, 226).
203 = ab0
203 = a
y = 203bx
226 = 203b10
1.113300493 = b10
b = 1.1133004931/10
b = 1.010790706
y = 203(1.010790706)x
Assume the population of the US is increasing exponentially with time. The
1970 census showed that the population was about 203 million. The 1980
census showed that the population had grown to about 226 million.
y = 203(1.010790706)x
3. Use your equation to predict the population this year.
2005 - 1970 = 35
y = 203(1.010790706)35
y = 295.55577
The model predicts there are about 295.6 million people in the
US this year.
Assume the population of the US is increasing exponentially with time. The
1970 census showed that the population was about 203 million. The 1980
census showed that the population had grown to about 226 million.
y = 203(1.010790706)x
4. Predict the year in which the population will reach 400 million.
400 = 203(1.010790706)x
1.97044335 = 1.010790706x
log 1.97044335
x»
log 1.010790706
x » 63.19433147
1970 + 64 = 2034
The population will reach 400 million in the year 2034.
Assume that the trade in value of a smart phone decays
exponentially with time.
1. You have had your phone for 7 months, and it
has a trade in value of $280. When your phone
was 3 months old, its trade-in value was $330.
a) How much will the phone be worth after 18 months?
b) What was the phones trade in value when it was new?
First we’ll create a model of the value of the phone as a
function of the number of months since purchase. To do this
we’ll create a system of 2 exponential equations:
Use the general format y = abx
Make one equation for each data point (3, 330) and (7, 280)
280 = ab
3
330 = ab
7
330 = ab3
280 = ab7
Now we will divide one equation by the other
to combine them into one equation with one
variable
y = a0.9598
x
.8485 = b
.8485 =b
1/4
4
4(1/4)
b = 0.9598
y = a0.9598
x
Evaluate for a using either point.
330 = a0.9598
3
330
=a
3
0.9598
A =$ 373.23
y = (373.23)0.9598
x
y = (373.23)0.9598
Y = $178.33
18
Carbon-14 (14C or radiocarbon) has a half life of 5,730 years.
1. Determine the equation for the percent of 14C remaining in the
original sample after x half-life periods.
Let x = no. of half-life periods
Let y = percent of carbon remaining.
a = 100 (representing 100%)
b = 1/2
()
y=a b
x
æ 1ö
y = 100 ç ÷
è 2ø
x
Carbon-14 (14C or radiocarbon) has a half life of 5,730 years.
2. If you start with 4 g of 14C, how much will remain in the original
sample after 5 half-life periods?
Let x = no. of half-life periods
Let y = amount of carbon remaining.
a=4
b = 1/2
()
y=a b
x
æ 1ö
y = 4ç ÷
è 2ø
(
5
)
y = 4 0.03125 = 0.125
0.125 grams of 14C are remaining after 5 half-life periods.
Carbon-14 (14C or radiocarbon) has a half life of 5,730 years.
3. If you start with 4 g of 14C, how much will remain in the original
sample after 20,000 years?
Let x = no. of years since the 14C started decaying.
Let y = amount of carbon remaining.
a=4
b=?
()
2 = 4 (b )
y=a b
(.50)
x
()
1
1
éë 2 ùû = é 4 b
4
4 êë
.50 = b
(
= b
)
1
5730 5730
y » 0.3559
b » 0.9998790392
5730
5730
1
5730
5730
ù
úû
(
)
(
)
y » 4 0.9998790392
y » 4 0.9998790392
x
20000
Approximately
0.36 grams
remains after
20,000 years.
Carbon-14 (14C or radiocarbon) has a half life of 5,730 years.
3. If you start with 4 g of 14C, how much will remain in the original
sample after 20,000 years?
Let x = no. of years since the 14C started decaying.
Let y = amount of carbon remaining.
a=4
b = 1/2
()
y=a b
æ 1ö
y » 4ç ÷
è 2ø
x
æ 1ö
y = 4ç ÷
è 2ø
æ 1ö
y = 4ç ÷
è 2ø
x
5730
20,000
5730
3.490401396
(
y » 4 0.0889783776
y » 0.3559
y » 0.3559
)
Approximately
0.36 grams
remains after
20,000 years.
A rule-of-thumb used by car dealers is that the trade-in value of a car
decreases by 30% each year.
1. Suppose you own a car whose trade-in value is presently
$2350. How much will it be worth 5 years from now?
y = abx
Decreasing by 30% means 1 - .3 = .7 is the growth factor.
If the car is presently worth $2350, then this is year zero, and a
point on the graph is (0, 2350).
Therefore a = 2350.
y = 2350(0.7)x
y = 2350(0.7)5
y = 394.9645
The car will be worth about $394.96.
A rule-of-thumb used by car dealers is that the trade-in value of a car
decreases by 30% each year.
y = 2350(0.7)x
2. If the car is presently 6 years old, what was its trade in value
when it was new.
y = 2350(0.7)-6
y = 19974.67042
The car could be traded in for $19,974.67 when it was new.
3. The car actually cost $24,170 when it was new. How do you
explain the difference between this amount and what your
model predicted?