Download Set 3 exponential_functions

Exponential Functions and
Their Graphs, Logs and
Natural Logs, Rational
Functions and their Graphs
The exponential function f with base a is
defined by
f(x) = ax
where a > 0, a  1, and x is any real number.
For instance,
f(x) = 3x and g(x) = 0.5x
are exponential functions.
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2
The Graph of f(x) = ax, a > 1
y
Range: (0, )
(0, 1)
x
Domain: (–, )
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Horizontal Asymptote
y=0
3
The Graph of f(x) = ax, 0 < a <1
y
Range: (0, )
Horizontal Asymptote
y=0
(0, 1)
x
Domain: (–, )
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4
Example: Sketch the graph of f(x) = 2x.
x
-2
-1
0
1
2
y
f(x) (x, f(x))
¼
½
1
2
4
(-2, ¼)
(-1, ½)
(0, 1)
(1, 2)
(2, 4)
4
2
x
–2
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2
5
Example: Sketch the graph of g(x) = 2x – 1. State the
domain and range.
The graph of this
function is a vertical
translation of the
graph of f(x) = 2x
down one unit .
y
f(x) = 2x
4
2
Domain: (–, )
x
Range: (–1, )
y = –1
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6
Example: Sketch the graph of g(x) = 2-x. State the
domain and range.
y
The graph of this
function is a
reflection the graph
of f(x) = 2x in the yaxis.
f(x) = 2x
4
Domain: (–, )
Range: (0, )
x
–2
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2
7
The irrational number e, where
e  2.718281828…
is used in applications involving growth and
decay.
Using techniques of calculus, it can be shown
that
n
 1
1    e as n  
 n
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8
The Graph of f(x) = ex
y
x
-2
-1
0
1
2
6
4
2
f(x)
0.14
0.38
1
2.72
7.39
x
–2
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2
9
Properties of Logarithmic Functions
If b, M, and N are positive real numbers,
b  1, and p and x are real numbers, then:
1. logb 1 = 0
Log15 1 = 0
1. logb 1 = 0
0
15 =
1. logb 1 = 0
2. logb b = 1
2. logb b = 1
Log
10
=
1
10
2. logb b = 1
1=
10
3. logb bx = x
3. logb bx = x
Log5 5x = x
3. logb bx = x
x
x = 5x
log
b
5
4. b
= x, x > 0
x
log
x = x, x > 0
4. b logb
b
log 3 x = x
4. b
= x, x > 0
3
5. logb MN = logb M + logb N
5. logb MN = logb M + logb N
5. log MN = log M + log N
1
10
Graph and find the domain of the following functions.
y = ln x
x
y
-2
-1
0
1
2
3
4
.5
cannot take
the ln of a (-)
number or 0
0
ln 2 = .693
ln 3 = 1.098
ln 4 = 1.386
ln .5 = -.693
D: x > 0
Graph y = 2x
y=x
x y
-2
-1
0
1
2
1
=
1
4
-1
2 =
2-2
1
2
4
2
The graph of y = log2 x
is the inverse of y = 2x.
The domain of y = b +/- loga (bx + c), a > 1 consists
of all x such that bx + c > 0, and the V.A. occurs when
bx + c = 0. The x-intercept occurs when bx + c = 1.
Ex.
Find all of the above for y = log3 (x – 2). Sketch.
D: x – 2 > 0
D: x > 2
V.A. @ x = 2
x-int. x – 2 = 1
x=3
(3,0)
3.6: Rational Functions and Their Graphs
Strategy for Graphing a Rational Function
Suppose that
p( x)
q( x)
where p(x) and q(x) are polynomial functions with no common factors.
f ( x) =
1. Determine whether the graph of f has symmetry.
f (-x) = f (x): y-axis symmetry
f (-x) = -f (x): origin symmetry
2. Find the y-intercept (if there is one) by evaluating f (0).
3. Find the x-intercepts (if there are any) by solving the equation p(x) = 0.
4. Find any vertical asymptote(s) by solving the equation q (x) = 0.
5. Find the horizontal asymptote (if there is one) using the rule for determining the
horizontal asymptote of a rational function.
6. Plot at least one point between and beyond each x-intercept and vertical asymptote.
7. Use the information obtained previously to graph the function between and beyond
the vertical asymptotes.
3.6: Rational Functions and Their Graphs
EXAMPLE: Graphing a Rational Function
Graph:
3x 2
f ( x) = 2
.
x -4
Solution
3  -x2 
3x 2
Step 1 Determine symmetry: f(-x) =
= f(x):
= 2
2
 -x  - 4
x -4
Symmetric with respect to the y-axis.
Step 2 Find the y-intercept:
3  02
0
f(0) = 2
= 0: y-intercept is 0.
=
0 - 4 -4
Step 3 Find the x-intercept:
3x2 = 0, so x = 0: x-intercept is 0.
Step 4 Find the vertical asymptotes: Set q(x) = 0.
x2 - 4 = 0 Set the denominator equal to zero.
x2 = 4
x = 2
Vertical asymptotes: x = -2 and x = 2.
more
3.6: Rational Functions and Their Graphs
EXAMPLE: Graphing a Rational Function
3x 2
f ( x) = 2
.
x -4
Graph:
Solution
Step 5 Find the horizontal asymptote:
y = 3/1 = 3.
Step 6 Plot points between and beyond the
x-intercept and the vertical asymptotes. With
an x-intercept at 0 and vertical asymptotes at x =
2 and x = -2, we evaluate the function at -3, -1,
1, 3, and 4.
x
-3
-1
1
3
4
7
6
5
4
3
Horizontal
asymptote: y = 3
-5
-4
27
27
3x 2
Vertical
4
f(x) = 2
-1 -1
asymptote: x = -2
5
x -4 5
The figure shows these points, the y-intercept,
the x-intercept, and the asymptotes.
x-intercept and
y-intercept
2
1
-3
-2
-1
1
-1
-2
-3
2
3
4
5
Vertical
asymptote: x = 2
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