Section 1.3 – FunctionsMath 1314 Page 1 of 13 Section 1.3 Section 1.3 – Functions Most of the...

Math 1314 of 13 Section 1.3

Section 1.3 – Functions

Most of the course will deal with functions. You’ll need to be able to evaluate functions, state the domain of a function, combine functions and read information from the graph of a function. First, here’s the definition of a function. A function f is a rule that assigns to each element in a set A one and only one element of a set B. The set A is called the domain of f. Sets A and B are sets of real numbers.

Functions are so important that we use a special notation when working with them. We’ll write ( )f x to denote the value of function f at x. We read this as “f of x.” The range of a function is all of the possible values of the function as x varies through the numbers in the domain. We can use letters other than f to denote a function, so you may see a function such as ( ), ( ) or ( )g x h x P x . For a function that comes from a word problem, we’ll usually choose a letter for the function that makes sense in the context of the problem, so a cost function will be ( )C x and a volume function will be ( )V x . Using Function Notation Example 1: Consider these four functions:

2 3 2

4

( ) 3 4, ( ) 5 2, ( ) 0.001 0.05 10.4 25000,

( ) 3 x

f x x g x x x h x x x x

j x e

= + = − + = − + + +=

Find:

A. ( 3)f − B. ( 1)j − C. (3000)h D. (2 5)f x − E. ( ) ( )g a b g a+ −

Solution: A. ( 3) 3( 3) 4 9 4 5f − = − + = − + = −


B. 4( 1) 44

3( 1) 3 3 0.0549j e e

e− −− = = = ≈

C. ( ) ( ) ( )3 2(3000) 0.001 3000 0.05 3000 10.4 3000 25000 26493800h = − + + + = −

D. ( ) ( )2 2 2(2 5) 2 5 5 2 5 2 4 20 25 10 25 2 4 30 52f x x x x x x x x− = − − − + = − + − + + = − +

E. 2 2 2 2( )( ) ( 5 2) 3( 5 2) 4 3 15 6 4 3 15 10f g x f x x x x x x x x= − + = − + + = − + + = − +� F. First compute ( ) :g a b+

( ) ( )2 2 2( ) 5 2 2 5 5 2g a b a b a b a ab b a b+ = + − + + = + + − − +

Then compute ( ) ( ) :g a b g a+ −

( )2 2 2

2 2 2

2

( ) ( ) 2 5 5 2 5 2

2 5 5 2 5 2

2 5

g a b g a a ab b a b a a

a ab b a b a a

ab b b

+ − = + + − − + − − +

= + + − − + − + −= + −

***

Example 2: Suppose 2 1, 1

( )3 1, 1

x xf x

x x

+ <=

− ≥. Find ( 3)f − and (5)f .

Solution: This is a piecewise defined function. The problem divides the domain of the function into two parts, those values less than 1 and those values greater than or equal to 1. Each part of the domain is assigned a function. You have to figure out which line applies to the given value of x, then use that line to evaluate the function. First, find ( 3)f − . Since 3 1− < , we’ll look at the top line.

( )2( 3) 3 1 9 1 10f − = − + = + =

Next, find (5)f . Since 5 1> , we’ll look at the bottom line.

( )(5) 3 5 1 15 1 14f = − = − =

***


Sometimes a function evaluation problem will look like a word problem. Example 3: A study of driving costs based on 2002 model compact cars found that the average cost measured in cents per mile is approximated by the function

2.2

2095( ) 32.08C x

x= + , where x (in thousands) denotes the number of miles the car

is driven in one year. (The average cost includes car payments, gas, insurance, maintenance and depreciation.) Find the average cost of driving a compact car 25000 miles in one year. Solution: We are just being asked to evaluate this function at a number, but we must be careful − the problem states that x is given in thousands of miles. Since we want to find the cost of driving 25,000 miles, we should substitute 25 for x.

2.2

2095(25) 32.08 33.84

25C = + = or approximately 34 cents per mile driven.

*** Most often, we’ll be interested in the graph of a function. It’s helpful to know what set of values will work in the function – that is, it’s helpful to know the domain. For many problems, stating the domain will be our first task. We’ll state domain using interval notation, so we’ll start with a review of that.

Interval Notation

Interval notation is a shorthand method for stating a set of numbers.

The interval (2, 5) means all real numbers between 2 and 5, not including either 2 or 5. If you want to include an endpoint, you use a bracket instead of a parenthese. So [2, 5) means all real numbers between 2 and 5, including 2 but not including 5. The interval [2, 5] includes both endpoints. You can use interval notation to indicate, say, all real numbers less than 3. You will write this as ( ), 3−∞ . This can also be written as 3x < . Similarly, all

numbers greater than or equal to 4 can be noted as [ )4,∞ . This can be written as

4x ≥ . You will most likely be asked to state answers in interval notation, so you should be able to rewrite the inequality using an interval.


Example 4: Write using interval notation: 6x ≥ . Solution: This inequality includes 6 and all real numbers bigger than 6. In interval notation: [ )6, ∞ .

***

Example 5: Write using interval notation: 3 12x− < ≤ . Solution: This inequality includes 12, does not include 3− and includes all real numbers between 3− and 12. In interval notation: ( ]3,12− .

***

Example 6: Write using interval notation: 0x ≠ . Solution: If 0x ≠ , then 0x < or 0x > . Write two intervals and connect them with the symbol, ∪ . We have ( ) ( ),0 0,−∞ ∪ ∞ .

***

Domain The domain of a function is the set of all x values that “work” in the function. We start by assuming that all real numbers will work, then check to see if we have to exclude any numbers from the domain. For example, given the function

2( )f x x= , there are no trouble spots. Any real number works in this function.

The domain is “all real numbers” which we will write as ( ), .−∞ ∞

The functions that require more analysis are radical functions, rational functions and logarithmic functions. We’ll look at examples of each.

Example 7: Find the domain: 2 5

( )3 6

xf x

x

+=−

.

This is a rational function. We have the binomial 3 6x − in the denominator. Since we cannot let the denominator be equal to zero, 3x – 6 cannot be equal to zero. So we have 3 6 0x − ≠ which we solve for x to find 2x ≠ .


This tells us that all real numbers except 2 will work. We will write the answer using interval notation, so the answer is( , 2) (2, )−∞ ∪ ∞ .

*** Example 8: Find the domain: ( ) 1f x x= + Solution: This is a radical function. The quantity under the square root sign must be non-negative (i.e., either 0 or positive), so we must have

1 0

1

x

x

+ ≥≥ −

.

In interval notation, [ )1,− ∞ .

*** Example 9: Find the domain: ( )( ) ln 5f x x= −

Solution: The domain of the function ( ) ln( )f x x= is ( )0,∞ . From this, we see

that 5x − must be positive. This is,

5 0

5

x

x

− >>

In interval notation, ( )5,∞ .

*** You will also need to be able to read information from the graph of a function. You should be able to find the points where the function crosses the x axis (x intercepts, or zeros of the functions), the point where the function crosses the y axis (y intercept), function values at a given value of x and equations of vertical and horizontal asymptotes. You should also be able to find the domain and the range of the function by looking at the graph of the function.


Example 10: Given the graph of ( ),f x find the domain, range, (2)f and (5).f

Solution: From the graph, we see that there are no x values less than 4− and there are no x values greater than 5. So the domain is [ ]5,4− . There are no y values below 5− and there are no y values above 4. So the range is [ ]5, 4− .

We can locate (2)f on the graph. Look along the x axis until you find the number 2. Then move up or down until your finger is on the graph of the function. Read off the corresponding y value. In this case, it’s 2− . So (2) 2f = − . Similarly, to find (5)f move your finger along the x axis until you find the number 5. Then move up or down until your finger is on the graph of the function. Read off the y value. In this case, it’s 5− . So (5) 5f = − .

***


Example 11: Look at the graph that is given below and state the equations of any vertical asymptotes and any horizontal asymptotes.

Solution: A vertical asymptote is a vertical line that the graph gets close to. In the graph shown above, there is a vertical asymptote when 1x = . A horizontal asymptote is a vertical line that the graph gets close to. In the graph shown above, there is a horizontal asymptote when 2.y =

***

Graphing

You should be able to tell by looking at a graph of an equation whether or not it is a function. Recall the vertical line test: if a vertical line can cross a graph in more than one place, the graph is not the graph of a function. Example 12: Determine which of these are graphs of functions: A. B.

Solution: A is not a function. At x = 1, for example, we have two y values. B is a function. It passes the vertical line test.

***


You should be able to graph certain functions quickly, without a table of values or looking at a calculator. Here are some examples. Example 13: Graph 2 3y x= + . Solution: From this equation, you know that the slope of the line is 2 and the y

intercept is 3. Graph the point (0, 3) and use the slope2 rise

1 run= to get another point

or two. Connect the dots. The graph on the right show the graph of the line.

*** You should be able to graph a basic function using transformations. We’ll use the function 2( )f x x= as an example. You should know what the graph of 2( )f x x= looks like.


You should be able to transform this – or any other basic function.

Vertical shifting: A vertical shift moves the function up or down. To graph ( ) , 0y f x c c= + > , start with the graph of ( )f x and shift it upward c units. To graph ( ) , 0y f x c c= − > , start with the graph of ( )f x and shift it downward c units. So 2( ) 5f x x= + moves the basic function up 5 units, while 2( ) 1f x x= − moves the basic function down one unit. Horizontal shifting: A horizontal shift moves the function left or right. To graph ( ), 0,y f x c c= + > start with the graph of ( )f x and shift it left c units. To graph ( ), 0,y f x c c= − > start with the graph of ( )f x and shift it right c units.

So ( )2( ) 7f x x= + moves the basic function 7 units to the left, while

( )2( ) 5f x x= − moves the basic function 5 units to the right.

Reflecting: A reflection of a function is its mirror image about the x axis or the y axis. To graph ( )f x− , reflect the graph of ( )f x about the x axis. To graph ( )f x− reflect the graph of ( )f x about the y axis. So 2( )f x x= − will be a reflection of the basic graph about the x axis and

( )2( )f x x= − will be a reflection of the basic graph about the y axis.

Vertical stretching or shrinking: A vertical stretch will move your graph closer to the y axis, while a vertical shrink will move it closer to the x axis. To graph ( ), 1y af x a= > , stretch the graph of ( )f x by a factor of a.

To graph ( ), 0 1y af x a= < < , shrink the graph of ( )f x by a factor of 1

a.


Example 14: Graph each.

A. 2( ) 1f x x= +

B. ( )2( ) 2f x x= −

C. 2( ) 3f x x= − +

D. 2( ) 2f x x=

E. 21( )

2f x x=

Solution: A. The vertex of the basic graph is ( )0,0 . Move it up one unit to ( )0,1

and sketch the graph.

B. Move the vertex of the basic graph to the right two units and then sketch the graph. The vertex will be at the point ( )2,0 .


C. This function involves two transformations, a reflection about the x axis and a vertical shift up three units. When a problem has more than one transformation, you’ll need to take care to perform the transformations in a correct order, to make sure that you draw the correct graph. Start by listing the transformations that should be done, and then perform them based on this order: (1) reflect over the x axis; (2) vertically stretch or shrink the graph; (3) translate the graph and then (4) reflect over the y axis. Since we have a reflection over the x axis and a vertical shift, we’ll do them in that order (starting with (1) and ending with (3) from the list given).

Try this in the opposite order, and you’ll see that you get a different – and incorrect – graph. D. The graph of this function has a vertical stretch by a factor of 2. Each function value from the basic function is multiplied by 2 to obtain 2( ) 2f x x= . The affect this has on the graph is to make to “narrower” – or closer to the y axis.


E. In this example, the graph of the function has a vertical shrink, since the number out front is less than 1. This has the effect of making the graph “wider” – or closer to the x axis.

*** In the next example, we’ll need to look at the graph of a function and see what we can read from it. Example 15: What do you know about the equation whose graph is given?

Solution: We start by observing that this is the graph of a rational function, since the graph of the function has two vertical asymptotes. These asymptotes must come from the denominator of the function. The vertical asymptotes are at 1x = and 1x = − , so we have factors of ( )1x − and ( )1x + in the denominator of the

rational function. We have two x intercepts, 2− and 2. This tells us that the function has factors of ( )2x − and ( )2x + in the numerator.


We can also see a horizontal asymptote at y = 1. This tells us that the function has the same highest power of x in the numerator as it has in the denominator, and that the ratio of their coefficients is 1, so they are the same. We have a y intercept at ( )0,4 .

So a good guess of the function is( )( )( )( )

2

2

2 2 4( )

1 1 1

x x xf x

x x x

− + −= =− + −

. This function

meets all the criteria we noted above.

***

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