Download exponential function

y = x2
y = 2x
The function f(x) = bx is an exponential function with
base b, where b is a positive real number other than 1
and x is any real number.
An asymptote is a line that a graph approaches (but
does not reach) as its x- or y-values become very
large or very small.
x
1
x
Graph y1 = 2 and y2 =
2
When b > 1, the function f(x) = bx represents
exponential growth.
When 0 < b < 1, the function f(x) = bx represents
exponential decay.
Graph f(x) = 2x along with each function below. Tell
whether each function represents exponential growth or
exponential decay. Then give the y-intercept.
a) 4  f(x)
y = 4(2x)
exponential growth, since the base, 2, is > 1
y-intercept is 4 because the graph of f(x) = 2x, which has
a y-intercept of 1, is stretched by a factor of 4
x
1
x
b) 6  f( x) y  6  2  6   
2
exponential decay, since the base, ½, is < 1
y-intercept is 6 because the graph of f(x) = 2x, which has
a y-intercept of 1, is stretched by a factor of 6
The total amount of an investment, A, earning compound
interest is
nt
r
A(t)  P  1   ,
n

where P is the principal, r is the annual interest rate,
n is the number of times interest is compounded per year,
and t is the time in years.
Find the final amount of a $500 investment after 8 years at
7% interest compounded annually, quarterly, and monthly.
nt
r

A(t)  P  1  
n

18
0.07 

A(t)

500
1

compounded annually:
= $859.09


1 

4 8
0.07 

compounded quarterly: A(t)  500  1 
= $871.11

4 

128
0.07 

compounded monthly: A(t)  500  1 

12


= $873.91
Find the final amount of a $2200 investment at 9% interest
compounded monthly for 3 years.
1. The domain of f (x) = bx consists of all real numbers. The range
of f (x) = bx consists of all positive real numbers.
2. The graphs of all exponential functions pass through the point (0,
1) because f (0) = b0 = 1.
3. If b > 1, f (x) = bx has a graph that goes up to the right and is an
increasing function.
4. If 0 < b < 1, f (x) = bx has a graph that goes down to the right and
is a decreasing function.
5. f (x) = bx is a one-to-one function and has an inverse that is a
function.
6. The graph of f (x) = bx approaches but does not cross the x-axis.
The x-axis is a horizontal asymptote.
f (x) = bx
f (x) = bx
0<b<1
b>1
Transformation
Equation
Description
Horizontal
translation
g(x) = bx+c
• Shifts the graph of f (x) = bx to the left c units if c > 0.
Vertical
stretching or
shrinking
g(x) = c bx
Multiplying y-coordintates of f (x) = bx by c,
• Stretches the graph of f (x) = bx if c > 1.
• Shrinks the graph of f (x) = bx if 0 < c < 1.
Reflecting
g(x) = -bx
g(x) = b-x
• Reflects the graph of f (x) = bx about the x-axis.
g(x) = -bx + c
• Shifts the graph of f (x) = bx upward c units if c > 0.
Vertical
translation
• Shifts the graph of f (x) = bx to the right c units if c < 0.
• Reflects the graph of f (x) = bx about the y-axis.
• Shifts the graph of f (x) = bx downward c units if c < 0.
Use the graph of f (x) = 3x to obtain the graph of g(x) = 3 x+1.
Solution Examine the table below. Note that the function g(x) = 3x+1 has
the general form g(x) = bx+c, where c = 1. Because c > 0, we graph g(x) = 3
x+1 by shifting the graph of f (x) = 3x one unit to the left. We construct a table
showing some of the coordinates for f and g to build their graphs.
f (x) = 3x
g(x) = 3x+1
(-1, 1)
-5
-4
-3
-2
-1
(0, 1)
1
2
3
4
5
6
Sketch a graph using transformation of the following:
1.
f ( x)  2  3
2.
f ( x)  2 x  1
3.
x
f ( x)  4
x 1
1
Recall the order of shifting: horizontal, reflection (horz.,
vert.), vertical.
An irrational number, symbolized by the letter e, appears as the base in
many applied exponential functions. This irrational number is approximately
equal to 2.72. More accurately,
The number e is called the natural base. The function f (x) = ex is called the
natural exponential function.
f (x) = 3x
f (x) = ex
e
2.71828...
4
f (x) = 2x
(1, 3)
3
(1, e)
2
(1, 2)
(0, 1)
1
-1
For continuous compounding:
A=
rt
Pe
Must pass the
horizontal line
test.
Is this function
one to one?
Yes
Does it have
an inverse?
Yes
Definition:
Logarithmic function of base “a” For x > 0, a > 0, and a  1,
y = logax if and only if x = ay
Read as “log
base a of x”
f(x) = logax is called the logarithmic
function of base a.
The most important
thing to remember
about logarithms is…
a logarithm is
an exponent.
log381 = 4
34 = 81
log168 = 3/4
163/4 = 8
Write the exponential equation in
logarithmic form
82 = 64
log 8 64 = 2
4-3 = 1/64
log4 (1/64) = -3
f(x) = log232
f(x) = log42
4y = 2
22y = 21
y = 1/2
f(x) = log31
3y = 1
y=0
Think:
y = log232
Step 1- rewrite it as an
exponential equation.
y = 132
2
f(x) = log10( /100)
Step 2- make the bases the
10ysame.
= 1/100
10y = 2
10y-2= 25
y = -2 Therefore,
y=5
You can only use a
calculator when the
base is
10
Find the log key on your calculator.
Evaluate the following using that log key.
log 10 = 1
log 1/3 = -.4771
log 2.5 = .3979
log -2 = ERROR!!!
Why?
 loga1
0
a
= 0 because = 1
 logaa = 1 because a1 = a
 logaax = x and alogax = x
 If logax = logay, then x = y
Rewrite as an exponent
4y = 1
Therefore, y = 0
log41= 0
log77 = 1
log620
6
Rewrite as an exponent
y
7 = 7
Therefore, y = 1
= 20
log3x = log312
x = 12
log3(2x + 1) = log3x
2x + 1 = x
x = -1
log4(x2 - 6) = log4 10
x2 - 6 = 10
x2 = 16
x = 4
Review:
How do you find the inverse of a
function?
Application of what you
know…
What is the inverse of f(x) = 3x?
y = 3x
Rewrite the
x = 3y
exponential as a
y = log3x
logarithm…
f-1(x) = log3x
Find the inverse of the following
exponential functions…
f(x) = 2x
f(x) = 2x+1
f(x) = 3x- 1
f-1(x) = log2x
f-1(x) = log2x - 1
f-1(x) = log3(x + 1)
Find the inverse of the following
logarithmic functions…
f(x) = log4x
f(x) = log2(x - 3)
f(x) = log3x – 6
f-1(x) = 4x
f-1(x) = 2x + 3
f-1(x) = 3x+6
Graphs of Logarithmic Functions
It is the inverse of y = 3x



x
y
=
3
y= log3x
Therefore,
the

Domain?
table of values
for (0,)
x
y
x
y
g(x) will-1be the
1/
1/
-1
3
Range?3 (-,)
reverse of the
0
1
1
0
table of values
for
Asymptotes?
x=0
3
1
y = 31x. 3
2
9
9
2
Graphs of Logarithmic Functions



g(x) = log4(x – 3)
What is the inverse
exponential function?
y= 4x + 3
Show your tables of
values.
y= log4(x – 3)
y= 4x + 3
Domain?
(3,)
x
y
x
y
3.25
-1
-1Range?
3.25(-,)
4
0
0
4
Asymptotes? x = 3
7
1
1
7
19
2
2
19
Graphs of Logarithmic Functions



g(x) = log5(x – 1) + 4
What is the inverse
exponential function?
y= 5x-4 + 1
Show your tables of
values.
y= 5x-4 + 1
y= log5(x – 1) + 4
x
xDomain?
y (1,)
1.2
3Range?
1.2 (-,)
2
4
2
Asymptotes? x = 1
6
5
6
26
6
26
y
3
4
5
6
The function defined by
f(x) = logex = ln x, x > 0
is called the natural logarithmic
function.
Find the ln
key on your calculator.
Evaluate the following using that ln key.
ln 2
=
.6931
ln 7/8
=
-.1335
ln 10.3 =
ln -1
2.3321
= ERROR!!!
Why?
 ln1
= 0 because e0 = 1
 Ln e = 1 because e1 = e
 ln ex = x and eln x = x
 If ln x = ln y, then x = y
ln 1/e= -1
2 ln e = 2
5
ln
e =
5
Rewrite as an exponent
ey = 1/e
ey = e-1
Therefore, y = -1
Rewrite as an exponent
ln e = y/2
e y/2 = e1
Therefore, y/2 = 1 and
y = 2.
Graphs of Natural Log Functions
g(x) = ln(x + 2)
Show your table of
values.
y= ln(x + 2)



x
y
-2
error
Domain? (-2,)
-1
0
0 (-,)
.693
Range?
1
1.099
Asymptotes? x = -2
2
1.386
Graphs of Natural Log Functions
g(x) = ln(2 - x)
Show your table of
values.
y= ln(2 - x)



x
y
2
error
Domain? (-2,)
1
0
0 (-,)
.693
Range?
-1
1.099
Asymptotes? x = -2
-2
1.386