Download ( ) Exponential Functions — 5.2 & 5.3

Math 4 Pre-Calculus
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Exponential Functions — 5.2 & 5.3
Exponential Function
f ( x ) = ax ,
a > 0 and a ≠ 1
Domain:
Horizontal Asymptote at y = 0 . The horizontal asymptote will move up or down depending on a horizontal shift.
The range depends on the global behavior which can be determined by the horizontal asymptote and the sign of vertical
stretch/compression factor, denoted by the letter b in your book..
1

All “basic” exponential functions contain the points ( 0 , 1) , (1 , a ) and  − 1 ,  .
a

Exponential Growth Functions
f ( x ) = ax ,
a >1
Domain:
Graph:
If a > 1 , the graph is a “
shape : right side up” . Range:
In mathematical terms, this means: as x → ∞, f ( x ) → ∞
and as x → − ∞, f ( x ) → 0 +
Exponential Decay Functions
f ( x ) = ax ,
0< a <1
Domain:
Graph:
If 0 < a < 1 , the graph is a “
shape : left side up” . Range:
In mathematical terms, this means: as x → ∞, f ( x ) → 0+ and as x → − ∞, f ( x ) → ∞
1.
Graph the equation.
f ( x ) = 3x
Domain:
Value of base:
Horizontal Asymptote:
Range:
Points:
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
2.
Graph the equation.
f ( x) = −3
x
−2
Domain:
Value of base:
Horizontal Asymptote:
Transformations:
Range:
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
-4
-5
-6
-7
-8
Points:
3.
Graph the equation.
1
f ( x) =  
2
x −1
Domain:
Value of base:
Horizontal Asymptote:
Transformations:
Range:
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
-4
-5
-6
-7
-8
Points:
4.
Graph the equation.
1
f ( x) =  
2
−x
+3
Domain:
Value of base:
Horizontal Asymptote:
Transformations:
Range:
Points:
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
-4
-5
-6
-7
-8
Many applications of exponential functions rely on a certain irrational base. Let’s use the compound interest formula to fill in
the following table, if $1 is invested at an annual interest rate of 100% for 1 year for the various numbers of compounding
periods per year.
Compound Interest Formula
r

A = P 1 + 
n

5.
P = principal
r = annual interest rate expressed as a decimal
n = number of interest periods per year
t = number of years P is invested
A = amount after t years
nt
,
where
Compound Interest
If $5,000 is invested at a rate of 6% per year compounded quarterly, find the principal (amount in the account) after
a)
3 months
b) 3 years
n
A = 1 (1 +
1
n
)
n ⋅1
or A = (1 +
1
(1 + 1)1
=
4
( 1 + 1 4 )4
=
12
(1 + 11 2 )1 2
=
56
(1 + 15 6 )
56
=
365
=
1,0 0 0
=
1 0,0 0 0
=
1 0 0,0 0 0
=
1,0 0 0,0 0 0
=
365
1,000
10,000
100,000
1,000,000
(1 + 13 6 5 )
(1 + 11,0 0 0 )
(1 + 11 0,0 0 0 )
(1 + 11 0 0,0 0 0 )
(1 + 11,0 0 0,0 0 0 )
What can you conclude about the value of the expression ( 1 +
In mathematical terms, as n → ∞ , (1 +
1
n
)
n
→
1
n
1
)
n
n
)
n
as n gets really large?
The Number e
If n is a positive integer, then as n → ∞ , (1 +
In calculus, this is written as l i m (1 +
1
n →∞
n
)
n
1
n
)
n
→ e ≈ 2.71828 .
= e ≈ 2.71828
Continuously Compounded Interest Formula
A = P er t
6.
,
where
P = principal
r = annual interest rate expressed as a decimal
t = number of years P is invested
A = amount after t years
Continuously Compounded Interest
How much money, invested at an interest rate of 4% per year compounded continuously, will amount to $25,000 in 3 years?
Natural Exponential Function
f ( x ) = ex
( Note: e ≈ 2 . 7 1 8 2 8 )
Domain:
Horizontal Asymptote at y = 0 .
Range:
The natural exponential function contains
1

the points ( 0 , 1) , ( 1 , e ) and  − 1 ,  .
e

7.
Graph the equation.
f ( x ) = e− x + 2
Domain:
Transformations:
Points:
Range:
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
Find an exponential function of the form f ( x ) = b a x or f ( x ) = b a x + c that has the given graph or conditions.
8.
y-intercept ( 0 , 4 ) ; passes through point P ( 2 , 1 6 ) and has a horizontal asymptote at y = − 1
10
9.
9
8
7
6
5
4
3
2
1
0
-5 -4 -3 -2 -1-1 0
-2
1
2
3
4
5
Exponential Functions Are One-to-One
The exponential function f given by f ( x ) = a x for 0 < a < 1 or a > 1 is one-to-one. Thus, the following equivalent
conditions are satisfied for real numbers x1 and x2
1.
If x1 ≠ x2 , then a x1 ≠ a x2 .
2.
If a x1 = a x2 , then x1 = x2
Solve each equation.
2
10. 4 x =
11. 2 5x ⋅
( 12 )
( 15 )
3x
4−2x
( )
= 5x
3
⋅125
12. e− x e2 =
( 1e )
4 x +1 2
Exponential Decay Applications
The half-life of a substance (or population) with an exponential decay is a defining characteristic. The half-life of a substance
(or population) is the time it takes for one-half of the original amount in a given sample to decay. The half-life of an isotope
distinguishes one isotope from another and is often used in carbon-dating of an object.
13. Exponential Decay
Let Q represent a mass of carbon 14 (in grams), whose half-life is 5730 years. The quantity of carbon 14 present after t
years is given by Q ( t ) = 1 0
( 12 )
t
5730
.
a)
Determine the initial quantity of carbon 14 (when t = 0 ).
b)
Predict the amount of carbon 14 present after 2000 years.
c)
Sketch a graph of Q ( t ) at any time from t = 0 to t = 1 0, 0 0 0 .
Exponential Growth Applications
The doubling-time of a substance (or population) with an exponential growth is a defining characteristic. The doubling-time
of a substance (or population) is the time it takes for the substance (or population) to double its original amount.
14. Exponential Growth
A certain type of bacterium increases according to the model P ( t ) = 1 0 0
a)
Determine the initial amount of bacterium.
b)
Predict the amount of bacterium after 5 hours.
( 54 ) , where t
t
is the time in hours.
c)
Sketch a graph of P ( t ) at any time from t = 0 to t = 1 0 .
d)
Use the above graph to estimate the doubling-time of P ( t ) .
15. Exponential Modeling
In a research experiment, a population of fruit flies is increasing at an exponential growth rate. Initially there are 30 fruit
flies. After 2 days, there are 100 flies.
a)
Find a simple exponential function y = b a x that models the population of the fruit flies after x days.
b) Use the above model to predict the number of fruit flies after 10 days.
Law of Exponential Growth (or Decay) Formula
Let q0 be the value of a quantity q at time (that is, q0 is the initial amount of q ). If q changes instantaneously at a rate
proportional to its current value, then
q = q ( t ) = q0 er t ,
where r > 0 is a rate of growth (or r < 0 is the rate of decay) of q .
Note: The above formula is the “same” as the continuously compounded interest formula.
16. Population Growth Rate
The 1985 population estimate for India was 762 million, and the population has been growing continuously at a rate of
about 2.2% per year. Assume that this rapid growth rate continues.
a)
Write an exponential function to model India’s population as a function of years since 1985.
b)
Use the model above to predict India’s population in 2010.