Transcript Lesson 6-2 Exponential Functions

Objective: To identify and solve exponential
functions
Exponential Functions have the form:
f(x) = bx
where b >0 & b ≠ 1
A constant raised to a variable power.
Numb3rs – season 1 episode 8 identity crisis
(~10 min)
Enter into calculator and graph:
f(x) = 2x
[2nd] [table] to see list of data points
Notice how quickly exponential
functions grow.
If bu = bv then u = v
When b > 0 and b≠ 1
The graph of f(x) = bx always passes through the
points (0,1) and (1,b)
The graph of f(x) = bx is the reflection
about the
x
y-axis of the graph of f(x)=  1   b  x
b
x
The graph of f(x) = b has the horizontal
asymptote y = 0.
The domain of f(x) = bx is the set of real
numbers: the range is the set of positive
numbers.
f(x) = bx is increasing if b > 1; f(x) = bx is
decreasing if 0< b< 1.
f(x) = bx is a one-to-one function since it passes
the horizontal line test.
f(x) = 4x
x
-2
-1
y
1/16
0
1
2
3
1
4
16
64
¼
310 = 35x
10 = 5x
x=2
27 = (x-1)7 if x > 1
2 = x-1
3=x
33x = 9x-1
33x = 32(x-1)
3x = 2x -2
x = -2
28 = 2x+1
42x+1 = 411
8x+1 = 2
e is an irrational number
It is 1  m1  , as m gets larger and larger.
It is approximately 2.71828
m
f(x)= ex is a natural exponential function
graph f(x)= ex on the calculator
Solve for x:
1. 53 − 2x =5−x
2. 32a=3−a
3. 31 − 2x= 243
Compound Interest
Continuous Compounding
Exponential Growth or decay (bacteria/
radiation half life)
Compound interest means the each payment is
calculated by including the interest previously
earned on the investment.
Year
Investment at Start
Interest
Investment at End
0 (Now)
$1,000.00
($1,000.00 × 10% = )
$100.00
$1,100.00
1
$1,100.00
($1,100.00 × 10% = )
$110.00
$1,210.00
2
$1,210.00
($1,210.00 × 10% = )
$121.00
$1,331.00
3
$1,331.00
($1,331.00 × 10% = )
$133.10
$1,464.10
4
$1,464.10
($1,464.10 × 10% = )
$146.41
$1,610.51
5
$1,610.51
If you have a bank account whose principal
= $1000, and your bank compounds the
interest twice a year at an interest rate of
5%, how much money do you have in your
account at the year's end?
When n gets very large it approaches becoming
continuous compounding. The formula is:
A  Pe
rt
P = principal amount (initial investment)
r = annual interest rate (as a decimal)
t = number of years
A = amount after time t
An amount of $2,340.00 is deposited in a bank
paying an annual interest rate of 3.1%,
compounded continuously. Find the balance
after 3 years.
Solution
A = 2340 e(.031)(3)
A = 2568.06
A = Pert ...or... A = Pekt ...or... Q =ekt
...or... Q = Q0ekt
k is the growth constant
In t hours the number of bacteria in a culture
will grow to be approximately Q = Q0e2t where
Q0 is the original number of bacteria. At 1 PM
the culture has 50 bacteria. How many bacteria
does it have at 4 PM? at noon?
Q = 50e2(3)
Q = 50e6
Q = 20,248
Q = 50e2(-1)
Q = 50e-2
Q=7
1. If you start a bank account with $10,000 and
your bank compounds the interest quarterly at
an interest rate of 8%, how much money do you
have at the year’s end ? (assume that you do
not add or withdraw any money from the
account)
2. An amount of $1,240.00 is deposited in a
bank paying an annual interest rate of 2.85 %,
compounded continuously. Find the balance
after 2½ years.
1.
2. A = 1240e(.0285)(2.5)
= $1,331.57
An artifact originally had 12 grams of carbon-14
present. The decay model A = 12e-0.000121t
describes the amount of carbon-14 present
after t years. How many grams of carbon-14
will be present in this artifact after 10,000
years?
A = 12e-0.000121t
A = 12e-0.000121(10,000)
A = 12e-1.21
A = 3.58