2 - 1

Section 2.1

Properties of Functions

Definition of a Function

Examples

Practice 1

Find the domain and range for the function

Solution: The domain includes only those values of x satisfying

since the denominator cannot be zero.

Using the methods for solving a quadratic inequality produces

the domain

Because the numerator can never be zero, the denominator

can take on any positive real number except for 0, allowing y

to take on any positive value except for 0, so the range is

2

1.

4y

x

2 4 0,x

( , 2) (2, ).

(0, ).

Practice 2

Given the function

find each of the following.

(a)

(b) All values of x such that

(a) Solution: Replace x with the expression x + h and simplify.

(b) Solution: Set f (x) equal to − 5 and then add 5 to both sides to make one side equal to 0.

Continued

2( ) 2 3 4,f x x x

( )f x h( ) 5.f x

2( ) 2( ) 3( ) 4f x h x h x h 2 22( 2 ) 3( ) 4x xh h x h 2 22 4 2 3 3 4x xh h x h

Practice 2 continued

This equation does factor as

Set each factor equal to 0 and solve for x.

(2 1)( 1) 0.x x

1 0 or 2 1 0x x

11 or .

2x x

2

2

2

2 3 4 5

2 3 4 5 0

2 3 1 0

x x

x x

x x

2

The function is defined as

if < 0

2 if = 0

2 if > 0

(a) Find (-2), (0), and (3). (b) Determine the domain of .

(c) Graph .

f

x x

f x x

x x

f f f f

f

(d) Use the graph to find the range of .

(e) Is continuous on its domain?

f

f

Piecewise-defined Functions:Example:

Symmetry:

Even and Odd Functions

A function f is even if for every number x in its domain the number -x is also in its domain and

f(-x) = f(x)

A function f is odd if for every number x in its domain the number -x is also in its domain and

f(-x) = - f(x)

32h x x x

35 1g x x

23 24 xxxf

2 - 17

Section 2.2

Quadratic Functions; Translation

and Reflection

Translation and Reflection

Vertex location

Example

2 3 11( ) , and .

2 2y a x h k h k

Where: h= k=Rewrite in this form

Practice 1

For the function

(a) complete the square, (b) find the y-intercept, (c) find the x

intercepts, (d) find the vertex, and (e) sketch the graph.

Solution (a): To begin, factor 2 from the x-terms so the

coefficient of x2 is 1:

Next, we make the expression inside the parentheses a perfect

square by adding the square of one-half of the coefficient of x,

Continued

22 6 1y x x

22( 3 ) 1y x x

2 9 92 3 1

4 2y x x

2

3 112

2 2y x

Practice 1 continued

Solution (b):The y-intercept (where x = 0) is − 1.

Solution (c): To find the x-intercepts, solve

Use the quadratic formula to verify that the x-intercepts are at

Solution (d): The function is now in the form

(3 11) / 2.

22 6 1 0.x x

2 3 11( ) , and .

2 2y a x h k h k

3 11Hence vertex, is , .

2 2

23 11

22 2

y x

Translation and Reflection

Translation and Reflection

2 - 29

Section 2.3

Polynomial and Rational Functions

Examples

Practice 1

Graph

Solution: Using the principles

of translation and reflection,

we recognize that this is

similar to the graph of

but reflected vertically

(because of the negative in

front of x6 ) and 64 units up.

6( ) 64 .f x x

6 ,y x

Examples

Practice: Identify degree and leading coefficient for each below

Rational Functions

Example: f(x)=1/x

Examples of graphs of rational functions:

Notice the domain!Asymptotes

2 - 40

Section 2.4

Exponential Functions

Examples:

3 9

Now use the property:

If 0, 1, an d , then .x ya a a a

x

x

x

y

More examples of exponentials:

Most common base for exponential is: e = 2.718281828

This number is called: Euler number. Occurs in natural phenomena and is used for financial modeling.

Your Turn 1

Solve

Solution: Since the bases must be the same, write 25 as 52 and

125 as 53.

/2 325 125 .x x

2 /2 3 3(5 ) (5 )x x

3 95 5 Multiply exponents. x x

3 9

Now use the property:

If 0, 1, an d , then .x ya a a a

x

x

x

y

2 9

9 / 2.

x

x

Application: Interests

If m is made larger and larger, ultimately we get to:

Your Turn 2

Find the interest earned on $4400 at 3.25% interest compounded

quarterly for 5 years.

Solution: Use the formula for compound interest with P = 4400,

r = 0.0325, m = 4, and t = 5.

The investment plus the interest is $5172.97. The interest

amounts to $5172.97 − $4400 = $772.97.

1tm

rA P

m

5(4)0.0325

4400 14

A

5172.97 Use a calcula tor.

Your Turn 3

Find the amount after 4 years if $800 is invested in an account

earning 3.15% compounded continuously.

Solution: In the formula for continuous compounding,

let P = 800, t = 4 and r = 0.0315 to get

or $907.43.

rtA Pe

0.0315(4)800A e

907.43,

2 - 47

Section 2.5

Logarithmic Functions

Example 1

Example 2

If all the following variable expressions represent positive

numbers, then for a > 0, a ≠ 1, the statements in (a)–(c) are

true.

Your Turn 1

Write the expression

as a sum, difference, or product of simpler logarithms.

Solution: Using the properties of logarithms,

2 3log ( / )a x y

2 3

2 3

log ( / )

log ( ) log ( )

a

a a

x y

x y

2log 3loga ax y

Your Turn 2

Evaluate

Solution: Using the change-of-base theorem for

logarithms with x = 50 and a = 3 gives

3log 50.

3log 50

ln50ln3

Use a calcu

3.

la

561

.

.

tor

Your Turn 3

Solve for x:

Solution:

This leads to two solutions: x = − 4 and x = 2. But notice that

x = − 4 is not a valid value for x in the original equation, since

the logarithm of a negative number is undefined.

The only solution is, therefore, x = 2.

2 2log log ( 2) 3.x x

2 2

2

log log ( 2) 3

log [ ( 2)] 3

x x

x x

3( 2) 2x x

2

2

2 8

2 8 0 Subtract 8 from both si des.

x x

x x

( 4)( 2) 0 Fa or. ctx x

Your Turn 4

Solve for x:

Solution: Taking natural logarithms on both sides gives

12 3 .x x

1ln 2 ln3x x

( 1) ln 2 ln3 ln ln rux x r u

(ln 2) ln 2 (ln3)x x

Subtract (ln3) and ln 2

(ln

2) (ln3) ln 2

f

rom bo

th sides.

x x x

Factor .

(ln 2 ln3) ln 2

ln 21.7095.

ln 2 ln3

x

x

x

2 - 59

Section 2.6

Applications: Growth and Decay;

Mathematics of Finance

Your Turn 1

Yeast in a sugar solution is growing at a rate such that 5 g

grows exponentially to 18 g after 16 hours. Find the growth

function, assuming exponential growth.

Solution: The values of y0 and k in the exponential growth

function y = y0 e kt must be found. Since y0 is the amount

present at time t = 0, y0 = 5. To find k, substitute y = 18, t = 16,

and y0 = 5 into the equation y = y0 e kt .

Now take natural logarithms on both sides and use the power

rule for logarithms and the fact that

Continued

(16)18 5 ke

ln 1.e

ln18 ln5 (16)k

Your Turn 1 continued

The exponential growth function is

where y is the number of grams of yeast present after t hours.

0.085 ,ty e

ln18 ln516

0.08

k

k

Recap: Graphs of Basic Functions