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Section 1.1: Functions

What is a Function?

The natural world is full of relationships between quantities that change.  When we see these relationships, it is natural for us to ask “If I know one quantity, can I then determine the other?”  This establishes the idea of an input quantity, or independent variable, and a corresponding output quantity, or dependent variable.  From this we get the notion of a functional relationship in which the output can be determined from the input.  

For some quantities, like height and age, there are certainly relationships between these quantities. Defining these relationships in a useful and meaningful way takes some consideration.  Given a specific person and any age, it is easy enough to determine their height, but if we tried to reverse that relationship and determine age from a given height, that would be problematic, since most people maintain the same height for many years.

Function

A function is a rule for a relationship between an input quantity and an output quantity in which:

each input determines exactly one output.

We say “the output is a function of the input.”

The inputs are often called the “independent” variable and the outputs are often called the “dependent” variable.


In the height and age example above, is height a function of age? Is age a function of height?

In the height and age example above, it would be correct to say that height is a function of age, since each age uniquely determines a height.  For example, on my 18th birthday, I had exactly one height of 69 inches.

However, age is not a function of height, since one height input might correspond with more than one output age. For example, for an input height of 70 inches, there is more than one output of age since I was 70 inches at the age of 20 and 21.


At a coffee shop, the menu consists of items and their prices.  Is price a function of the item? Is the item a function of the price?

We could say that price is a function of the item, since each input of an item has one output of a price corresponding to it.  We could not say that item is a function of price, since two items might have the same price.


In many classes the overall percentage you earn in the course corresponds to a decimal grade point. Is decimal grade a function of percentage? Is percentage a function of decimal grade?

For any percentage earned, there would be a decimal grade associated, so we could say that the decimal grade is a function of percentage. That is, if you input the percentage, your output would be a decimal grade. Percentage may or may not be a function of decimal grade, depending upon the teacher’s grading scheme. With some grading systems, there are a range of percentages that correspond to the same decimal grade.


Sometimes in a relationship each input corresponds to exactly one output, AND every output corresponds to exactly one input.  We call this kind of relationship a one-to-one function.

One-to-One Function

A one-to-one function is a function for which each output corresponds to exactly one input.


From Example 3, if each unique percentage corresponds to one unique decimal grade point and each unique decimal grade point corresponds to one unique percentage then it is a one-to-one function.

Let’s consider bank account information.

  1. Is your balance a function of your bank account number?
    (if you input a bank account number does it make sense that the output is your balance?)

  2. Is your bank account number a function of your balance?
    (if you input a balance does it make sense that the output is your bank account number?)


Function Notation

To simplify writing out expressions and equations involving functions, a simplified notation is often used.  We also use descriptive variables to help us remember the meaning of the quantities in the problem.

Rather than write “height is a function of age”, we could use the descriptive variable h to represent height and we could use the descriptive variable a to represent age.

“height is a function of age”

if we name the function f we write

“h is f of a”

or more simply

$h =f(a)$

we could instead name the function $h$ and write

$h(a)$

which is read “h of a”

Remember we can use any variable to name the function; the notation $h(a)$ shows us that $h$ depends on $a$. The value “$a$” must be put into the function “$h$” to get a result. Be careful - the parentheses indicate that age is input into the function (Note: do not confuse these parentheses with multiplication! It does not mean “h times a”).

Function Notation

The notation output=$f(\text{input})$ defines a function named $f$.  This would be read “output is $f$ of input”.

The parentheses in function notation do NOT indicate multiplication! The parentheses merely surround the input to the function named $f$.

f(a)=b f is the name of the function (a) is the input value b is the output value


Introduce function notation to represent a function that takes as input the name of a month, and gives as output the number of days in that month.

The number of days in a month is a function of the name of the month, so if we name the function $f$, we could write “$\text{days} =f(\text{months})$” or $d =f(m)$. If we simply name the function $d$, we could write $d(m)$.

For example, $d(\text{March})=31$, since March has 31 days. The notation $d(m)$ reminds us that the number of days, $d$ the output, is dependent on the name of the month, $m$ the input.


A function $N=f(y)$ gives the number of police officers, $N$, in a town in year $y$.  What does $f(2005)=300$ tell us?

When we read $f(2005)=300$, we see the input quantity is 2005, which is a value for the input quantity of the function, the year ($y$). The output value is 300, the number of police officers ($N$), a value for the output quantity. Remember $N=f(y)$.  This tells us that in the year 2005 there were 300 police officers in the town.


Tables as Functions

Functions can be represented in many ways:  with words (as we did in the last few examples), tables of values, graphs, or formulas.  Represented as a table, we are presented with a list of input and output values.  

In some cases, these values represent everything we know about the relationship, while in other cases the table is simply providing us a few select values from a more complete relationship.

Table 1: This table represents the input, number of the month (January=1, February=2, and so on) while the output is the number of days in that month. This represents everything we know about the months and days for a given year (that is not a leap year)

(input)
Month number,
m		 (output)
Days in month,
D
131
228
331
430
531
630
731
831
930
1031
1130
1231

Table 2: The table below defines a function $Q=g(n)$. Remember this notation tells us $g$ is the name of the function that takes the input $n$ and gives the output $Q$.

$n$$Q$
18
26
37
46
58

Table 3: This table represents the age of children in years and their corresponding heights. This represents just some of the data available for height and ages of children.

(input) $a$, age in years(output) $h$, height inches
540
542
644
747
850
952
1054


Which of these tables define a function (if any)? Are any of them one-to-one?

Input Output
2 1
5 3
8 6


Input Output
-3 5
0 1
4 5


Input Output
1 0
5 2
5 4

The first and second tables define functions.  In both, each input corresponds to exactly one output.  The third table does not define a function since the input value of 5 corresponds with two different output values.

Only the first table is one-to-one; it is both a function, and each output corresponds to exactly one input. Although table 2 is a function, because each input corresponds to exactly one output, each output does not correspond to exactly one input so this function is not one-to-one. Table 3 is not even a function and so we don’t even need to consider if it is a one-to-one function.


If each percentage earned translated to one letter grade, would this be a function? Is it one-to-one?


Solving and Evaluating Functions:

When we work with functions, there are two typical things we do: evaluate and solve.  Evaluating a function is what we do when we know an input and use the function to determine the corresponding output.  Evaluating will always produce one result, since each input of a function corresponds to exactly one output.  

Solving equations involving a function is what we do when we know an output, and use the function to determine the inputs that would produce that output.  Solving a function could produce more than one solution, since different inputs can produce the same output.

Functions: Solving vs. Evaluating

If we know an input and use the function to determine an output, we are evaluating the function.

If we want to know what input produces an output, we are solving a function.


Using the table shown, where $Q=g(n)$

  1. Evaluate $g(3)$

  2. Solve $g(n)$=6

$n$12345
$Q$86768

  1. Evaluating $g(3)$ (read: “$g$ of 3”) means that we need to determine the output value, $Q$, of the function $g$ given the input value of $n=3$.  Looking at the table, we see the output corresponding to $n=3$ is $Q=7$, allowing us to conclude $g(3)=7$.

  2. Solving $g(n)=6$ means we need to determine what input values, $n$, produce an output value of 6.  Looking at the table we see there are two solutions: $n=2$ and $n=4$.

    When we input 2 into the function $g$, our output is $Q = 6$

    When we input 4 into the function $g$, our output is also $Q=6$


Using the function in Example 7, evaluate $g$(4)


Graphs as Functions

Oftentimes a graph of a relationship can be used to define a function.  By convention, graphs are typically created with the input quantity along the horizontal axis and the output quantity along the vertical.

The most common graph has $y$ on the vertical axis and $x$ on the horizontal axis, and we say $y$ is a function of $x$, or $y=f(x)$ when the function is named $f$.

A graph of x (horizontal) and y (vertical) axis on a grid.


Which of these graphs defines a function $y=f(x)$? Which of these graphs defines a one-to-one function?

Graph of a smooth curve that increases to negative 1 comma 3, decreases to 1 comma negative 3, then increases.

Graph of a decrease straight line
Graph of a circle with radius 3 centered at the origin.


Looking at the three graphs above, the first two define a function $y=f(x)$, since for each input value along the horizontal axis there is exactly one output value corresponding, determined by the $y$-value of the graph.  The 3rd graph does not define a function $y=f(x)$ since some input values, such as $x$=2, correspond with more than one output value.

Graph 1 is not a one-to-one function. For example, the output value 3 has two corresponding input values, -1 and 2.3

Graph 2 is a one-to-one function; each input corresponds to exactly one output, and every output corresponds to exactly one input.

Graph 3 is not even a function so there is no reason to even check to see if it is a one-to-one function.


The vertical line test is a handy way to think about whether a graph defines the vertical output as a function of the horizontal input.  Imagine drawing vertical lines through the graph.  If any vertical line would cross the graph more than once, then the graph does not define only one vertical output for each horizontal input, and therefore the graph is not a function.

Vertical Line Test

If a vertical line touches the graph of an equation more than once, then the graph does not represent a function. In this case, we would say the graph "fails the vertical line test."

This is because the vertical line is showing a location where an input corresponds to more than one output.


Once you have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line crosses the graph more than once, then the graph does not define a one-to-one function.

Horizontal Line Test

If a horizontal line touches the graph of a function more than once, then the graph does not represent a one-to-one function. In this case, we would say the graph "fails the horizontal line test."

This is because the horizontal line is showing a location where an output corresponds to more than one input.


Evaluating a function using a graph requires taking the given input and using the graph to look up the corresponding output.  Solving a function equation using a graph requires taking the given output and looking on the graph to determine the corresponding input.


Given the graph of $f(x)$

  1. Evaluate $f(2)$

  2. Solve $f(x)=4$

A U-shaped graph passing through negative 1 comma 4, 0 comma 1, 1 comma 0, 2 comma 1, and 3 comma 4


  1. To evaluate $f(2)$, we find the input of $x = 2$ on the horizontal axis.  Moving up to the graph gives the point $(2, 1)$, giving an output of $y=1$.   $f(2)=1$.

  2. To solve $f(x)=4$, we find the value 4 on the vertical axis because if $f(x)=4$ then 4 is the output.  Moving horizontally across the graph gives two points with the output of 4: (-1,4) and (3,4).  These give the two solutions to $f(x)=4$:  $x = − 1$ or $x =3$

    This means $f(−1) = 4$ and $f(3) = 4$, or when the input is -1 or 3, the output is 4.


Notice that while the graph in the previous example is a function, finding two input values for the output value of 4 shows us that this function is not one-to-one.

Using the graph from example 9, evaluate $f$(1) and solve $f$($x$) = 1.


Formulas as Functions

When possible, it is very convenient to define relationships using formulas.  If it is possible to express the output as a formula involving the input quantity, then we can define a function.

Express the relationship $2n + 6p =12$ as a function $p =f(n)$ if possible.

To express the relationship in this form, we need to be able to write the relationship where $p$ is a function of $n$, which means writing it as $p$=[something involving $n$].  So, we’ll work on isolating the $p$ in the equation.

\[ \begin{array}{lc} 2n + 6p=12 & \text{Subtract } 2n \text{ from both sides} \\ 6p=12 - 2n & \text{Divide both sides by } 6 \text{ and simplify} \\ p=\frac{12 - 2n}{6} & \\ p=\frac{12}{6} - \frac{2n}{6} & \text{Simplify each term} \\ p=2 - \frac{1}{3}n & \\ \end{array} \] Having rewritten the formula as $p$=, we can now express $p$ as a function:

$$p=f(n)=2 - \frac{1}{3}n$$


It is important to note that not every relationship can be expressed as a function with a formula.

Note the important feature of an equation written as a function is that the output value can be determined directly from the input by doing evaluations - no further solving is required.  This allows the relationship to act as a magic box that takes an input, processes it, and returns an output.  Modern technology and computers rely on these functional relationships, since the evaluation of the function can be programmed into machines, whereas solving things is much more challenging.

Express the relationship $x^2+y^2=1$ as a function $y =f(x)$ if possible.

If we try to solve for $y$ in this equation:

$$y^2=1-x^2$$

$$y=\pm \sqrt{1 - x^{2}}$$

We end up with two outputs corresponding to the same input (the positive root and the negative root), so this relationship cannot be represented as a single function $y=f(x)$.


As with tables and graphs, it is common to evaluate and solve functions involving formulas.  Evaluating will require replacing the input variable in the formula with the value provided and calculating.  Solving will require replacing the output variable in the formula with the value provided, and solving for the input(s) that would produce that output.

Given the function $f(x)=x^2-3x+2$, evaluate each of the following.

  1. $f(2)$

  2. $f(-2)$

  3. $f(a)$

  4. $f(a+h)$

  5. $f(a+h)-f(a)$

In this example, we’re trying to get used to working with function notation algebraically. When we’re evaluating our function, it’s useful to think of the variable as just a big box where we’re going to dump our inputs:

$$f(x)=x^2-3x+2$$
becomes
$$f(\Box)=(\Box)^2 - 3(\Box)+2$$

So, when we’re evaluating our function, we’ll put identical copies of the input into each box.

  1. Starting with $f(\Box) = (\Box)^2 − 3(\Box) + 2$ we'll evaluate $f(2)$ by replacing each and every input box with a 2.
    $$f(2)=(2)^2 - 3(2)+2$$ $$=4-6+2$$ $$=-2+2$$ $$=0$$ So, $f(2)=0$.

  2. This time, we’ll replace each box with -2. This is where it becomes very useful to always put our inputs in a set of parentheses!
    $$f(−2) = (−2)^2 − 3(−2) + 2$$ $$ = 4 + 6 + 2 $$ $$= 12$$ That means $f(-2)=12$.

  3. To find $f(a)$, we’ll follow the same procedure as before: we’ll replace each input box with $a$. We just won’t be able to simplify much after that!
    $$f(a)=(a)^2 - 3(a)+2$$ $$=a^2-3a+2$$ So $f(a)=a^2-3a+2$.

  4. Now, we want to fill each of our input boxes with ($a$+$h$). We’ll have to be careful with our algebra after that!
    $$f(\Box) = (\Box)^2 − 3(\Box) + 2$$

    becomes
    $$f(a+h)=(a+h)^2-3(a+h)+2$$ Remember, that $(a+h)^2$ means $(a+h)(a+h)$, so we have a lot of distributing to do!
    $$=(a+h)(a+h)+(-3)(a+h)+2$$ $$=a^2+ah+ah+h^2+(-3a)+(-3h)+2$$ $$=a^2 +2ah + h^2 -3a - 3h + 2$$

  5. Finally we come to $f(a+h)-f(a)$. We already know that $f(a) = a^2 − 3a + 2$, and $f(a+h) = a^2 + 2ah + h^2 − 3a − 3h + 2$, so we’ll put those together to build our full answer:
    $$f(a+h) − f(a) = a^2 + 2ah + h^2 − 3a − 3h + 2 − (a^2−3a+2) $$ $$= a^2 + 2ah + h^2 − 3a − 3h + 2 − a^2 + 3a − 2 $$ $$= a^2 − a^2 − 3a + 3a + 2 − 2 + 2ah + h^2 − 3h $$ $$= 2ah + h^2 − 3h$$


Given the function $g(x) = 5 − x^2$, evaluate the following:

  1. $g(3)$

  2. $g(−3)$

  3. $g(a+h)$

  4. $g(a)$

  5. $g(a+h) − g(a)$


Given the function $k(t) = 3(t−4)$

  1. Evaluate $k(2)$

  2. Solve $k(t)=1$

  1. “Evaluate $k(2)$” means we want to find the output when the input is $t=2$. We plug in the input value 2 into the formula wherever we see the input variable $t$, then simplify
    $$k(2) = 3(2−4)$$ $$ = 3(−2)$$ $$ =  − 6$$ So $k(2) =  − 6$.

  2. “Solve $k(t)$=1” means we want to find the input $t$ that will give us an output of 1. We set the formula for $k(t)$ equal to 1, and solve for the input
    \[ \begin{array}{rclc} 1 & =& 3(t - 4) & \text{Set formula equal to 1} \\ 1 &=& 3t - 12 & \text{Distribute} \\ 1 + 12 &=& 3t - 12 + 12 & \text{Add 12 to both sides} \\ 13 &=& 3t & \text{Simplify} \\ \frac{13}{3} & =& \frac{3t}{3} & \text{Divide by 3 on both sides} \\ t &=&\frac{13}{3} & \text{Simplify} \end{array} \]


Given the function $h(p) = p^2 + 2p$

  1. Evaluate $h(4)$

  2. Solve $h(p)=3$

  1. To evaluate $h(4)$ we substitute the value 4 for the input variable $p$ in the given function.

    $$h(4) = (4)^2 + 2(4)$$ $$ = 16 + 8$$ $$ = 24$$

  2. To solve $h$($p$) = 3, we’ll set our formula equal to 3 and solve. That means we want to solve $p$^2 + 2$p$ = 3. That’s a quadratic equation! Let’s see if we can factor and use the Zero Product Property.
    \[ \begin{array}{rcll} p^{2} + 2p & =& 3 \\ p^{2} + 2p - 3 & =& 3 - 3 &\text{Subtract 3 from both sides} \\ p^{2} + 2p - 3 & =& 0& \\ (p - 1)(p + 3) &=& 0& \text{Factor} \end{array} \]

By the zero product property since $(p−1)(p+3) = 0$, either $p − 1 = 0$ or $p + 3 = 0$ (or both of them equal 0) and so we solve both equations for $p$, finding $p = − 3$ from the first equation and $p =1$ from the second equation.

This gives us the solution: $h(p)=3$ when $p =1$ or $p = − 3$. 

We found two solutions in this case, which tells us this function is not one-to-one.


Given the function $g(m) = 4 − (2+m)$

  1. Evaluate $g(5)$

  2. Solve $g(m)=2$


Graphing Functions from Formulas

There will be many times when we want to analyze various features of functions. Sometimes these features are easier to spot and understand if we can start with a graph of a function. For that reason, we want to work on the skill of creating graphs from function formulas.

Create a graph for the function $f(x) = 3x − 4$

To create a graph for this function, we want to describe a bunch of input/output relationships. To do this, we’ll arbitrarily choose a list of input values, say x$-values from -2 to 2. We’ll evaluate the function at each of these inputs to find the corresponding outputs, then plot the points that we’ve determined.

$x$ $f(x)$ Point
$−2$ $f(−2) = 3(−2) − 4 = −10$ $(−2, −10)$
$−1$ $f(−1) = 3(−1) − 4 = −7$ $(−1, −7)$
$0$ $f(0) = 3(0) − 4 = −4$ $(0,  − 4)$
$1$ $f(1) = 3(1) − 4 = −1$ $(1, −1)$
$2$ $f(2) = 3(2) − 4 = 2$ $(2, 2)$

We didn’t have to pick $x$-values from -2 to 2. We could have chose 1.5, or 0.9, or 57. We’re just making a choice that is a) fairly easy to calculate and b) creates enough points to get an idea of what the graph looks like. Currently, if we graph these points, we see the following:

points arranged in an upward-sloped straight line

These are only some of the points that form the function though. To represent the graph of the complete function, we connect these points, in this case with a straight line:

A line connecting the points (-2,-10), (-1,7), (0,-4), (1,-1), and (2,2).


A big part of College Algebra is being able to quickly understand functions and their features. We’ll start learning ways to recognize ways to identify a function’s expected shape and other features; for now, we’ll choose enough $x$-values to get an idea of the function’s shape and sketch the graph as best we can.

Create a graph for the function $f(x) = (x − 1)^2$

Again, we’ll choose a set of $x$-values to substitute in, so we can create a group of points to graph. We can again work with values from -2 to 2.

$x$ $f(x)$ Point
$−2$ $f(−2) = (−2 − 1)^2 = (−3)^2 = 9$ $(−2, 9)$
$−1$ $f(−1) = (−1 − 1)^2 = (−2)^2 = 4$ $(−1, 4)$
$0$ $f(0) = (0 − 1)^2 = (−1)^2 = 1$ $(0, 1)$
$1$ $f(1) = (1 − 1)^2 = 0^2 = 0$ $(1, 0)$
$2$ $f(2) = (2 − 1)^2 = 1^2 = 1$ $(2, 1)$

When we graph these points, we don’t see a perfectly straight line anymore. In fact, we see a curved shape like this:

A u-shaped function through the points (-2,9), (-1,4), (0,1), (1,0), and (2,1).


Graph the function $f(x)=\sqrt{x - 1}$. Be careful about the $x$-values you choose!


Important Topics of this Section

Definition of a function

Input (independent variable)

Output (dependent variable)

Definition of a one-to-one function

Function notation

Descriptive variables

Functions in words, tables, graphs & formulas

Vertical line test

Horizontal line test

Evaluating a function at a specific input value

Solving a function given a specific output value


Section 1.1 Exercises

Conceptual Questions

  1. What is a function?
  2. What does it mean for a relation to be one-to-one? Can something be one-to-one but not a function?
  3. How do we evaluate functions from tables? From graphs? From formulas?
  4. How do we solve functions from tables? From graphs? From formulas?
  5. What does the vertical line test determine? Why does it work?
  6. What does the horizontal line test determine? Why does it work?
  7. How can we sketch a graph of a function?

Practice Problems

  1. The amount of garbage, $G$, produced by a city with population $p$ is given by $G$ =$f$($p$). $G$ is measured in tons per week, and $p$ is measured in thousands of people.

    1. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function $f$.

    2. Explain the meaning of the statement $f$(5) = 2.

    1. (a) $f(40) = 13$, because the input 40 (in thousands of people) gives the output 13 (in tons of garbage)



  2. The number of cubic yards of dirt, $D$, needed to cover a garden with area $a$ square feet is given by $D$ = $g$($a$).

    1. A garden with area 5000 ft^2 requires 50 cubic yards of dirt. Express this information in terms of the function $g$.

    2. Explain the meaning of the statement $g$(100) = 1.

  3. Let $f(t)$ be the number of ducks in a lake $t$ years after 1990. Explain the meaning of each statement:

    1. $f(5) = 30$

    2. $f(10) = 40$

    1. (a) In 1995 (5 years after 1990) there were 30 ducks in the lake.
      (b) In 2000 (10 years after 1990) there were 40 ducks in the lake.



  1. Let $h(t)$ be the height above ground, in feet, of a rocket $t$ seconds after launching. Explain the meaning of each statement:

    1. $h(1) = 200$

    2. $h$(2) = 350

  1. Select all of the following graphs which represent $y$ as a function of $x$.

    1. a horizontal line

    2. a curved shape that starts at the bottom left, goes up, then down, then up to the right.
    3. a sideways curved shape that starts on the left, snakes right, then left, then up and right.

    1. an oscillating shape, that goes up and down

    2. a diagonal line
    3. a circle

    1. Graphs (a) (b) (d) and (e) represent y as a function of x because for every value of x there is only one value for y. Graphs (c) and (f) are not functions because they contain points that have more than one output for a given input, or values for x that have 2 or more values for y.


  2. Select all of the following graphs which represent $y$ as a function of $x$.

  1. a large v shape

  2. a curved shape that starts in the lower left, goes down, then up a lot, then down a little, then up to the right.
  3. a c-shaped curve

  1. a vertical line

  2. a diagonal line
  3. an ellipse (oval)

 

  1. Select all of the following tables which represent $y$ as a function of $x$.

  1. x 5 10 15
    y 3 8 14
  1. x 5 10 15
    y 3 8 8
  1. x 5 10 10
    y 3 8 14

  1. Tables (a) and (b) represent y as a function of x because for every value of x there is only one value for y. Table (c) is not a function because for the input x=10, there are two different outputs for y.

  1. Select all of the following tables which represent $y$ as a function of $x$.

  1. x 2 6 13
    y 3 10 10
  1. x 2 6 6
    y 3 10 14
  1. x 2 6 13
    y 3 10 14

  1. Select all of the following tables which represent y as a function of x.

  1. x y
    0-2
    31
    46
    89
    31
  1. x y
    -1-4
    23
    54
    87
    1211
  1. x y
    0-5
    31
    34
    98
    1613
  1. x y
    -1-4
    12
    42
    97
    1213

  1. Tables (a) (b) and (d) represent y as a function of x because for every value of x there is only one value for y. Table (c) is not a function because for the input x=3, there are two different outputs for y.


  1. Select all of the following tables which represent $y$ as a function of $x$.

  1. x y
    -4-2
    32
    64
    97
    1216
  1. x y
    -5-3
    21
    24
    79
    1110
  1. x y
    -1-3
    12
    54
    98
    12
  1. x y
    -1-5
    31
    51
    87
    1412

  1. Select all of the following tables which represent $y$ as a function of $x$ and are one-to-one.

  1. x3812
    y477
  1. x3812
    y4713
  1. x388
    y4713

  1. Table (b) represents y as a function of x and is one-to-one because there is a unique output for every input, and a unique input for every output. Table (a) is not one-to-one because two different inputs give the same output, and table (c) is not a function because there are two different outputs for the same input x=8.


  1. Select all of the following tables which represent y as a function of x and are one-to-one.

  1. x288
    y5613
  1. x2814
    y566
  1. x2814
    y5613

  1. Select all of the following graphs which are one-to-one functions.

  1. an ellipse

  2. a curved shape that starts on the left and curves upward
  3. a shape that starts almost flat on the left, and curves to the right and down steeply

  1. an oscillating shape, that goes up and down

  2. a diagonal line
  3. a shape that starts at the point (1,-2) and curves down and to the right

  1. Graphs (b) (c) (e) and (f) are one-to-one functions because there is a unique input for every output. Graph (a) is not a function, and graph (d) is not one-to-one because it contains points which have the same output for two different inputs.



  1. Select all of the following graphs which are one-to-one functions.

  1. a v shape

  2. a curved shape that starts at the bottom left, goes up, then down, then up to the right.
  3. a c shape

  1. a vertical line

  2. a diagonal line
  3. a shape in two parts. on the left, it is almost flat, then goes down drastically as x values get near 1. Then after x-values of 1, the graph is almost vertical, then goes down and to the right near the x-axis

 

Given each function $f(x)$ graphed, evaluate $f(1)$ and $f(3)$ and solve $f(x) = 1$.

  1. a curved shape through the points (-2,0), (-1, 3), (0,-2), (1,1), (2,3), and (3,0)

  1. a curved shape through the points (-1,-1), (0,1),(1,3),(2,1),(3,-1),(4,1),and (5,3).

  1. (a) $f(1) = 1$ (b) $f(3) = 1$


  1. Given the function $g$($x$) graphed here,

    1. Evaluate $g(2)$

    2. Solve $g(x) = 2$

    a curved shape through the points (-3,2),(0,3),(2,4).

  1. (a) $g(2) = 4$ (b) $g(−3) = 2$


  1. Given the function $f$($x$) graphed here.

    1. Evaluate $f(4)$

    2. Solve $f(x)=4$

a curved shape through the points (-3,4),(0,3),(4,2)


  1. Based on the table below,

  1. Evaluate $f(3)$

  2. Solve $f(x)=1$

x 0 1 2 3 4 5 6 7 8 9
f(x) 74 28 1 53 56 3 36 45 14 47

  1. (a) $f(3) = 53$ (b) $f(2) = 1$


  1. Based on the table below,

  1. Evaluate $f(8)$

  2. Solve $f(x)=7$

x 0 1 2 3 4 5 6 7 8 9
f(x) 62 8 7 38 86 73 70 39 75 34

For each of the following functions, evaluate: $f(−2), f(−1), f(0) f(0), f(1), f(2), f(a+h)$, and $f(a+h) − f(a)$.

  1. $f(x)=4-2x$

  2. $f(x)=8-3x$

  3. $f(x)=8 x^2-7 x+3$

  4. $f(x)=6 x^2-7 x+4$

  5. $f(x)=-x^3+2x$

  6. $f(x)=5x^4+x^2$

  7. $f(x)=3+\sqrt{x+3}$

  1. $f(x)=4-\sqrt[3]{x-2}$

  2. $f(x)=(x−2)(x+3)$

  3. $f(x)=(x+3)(x−1)^2$

  4. $f(x)=\frac{x-3}{x+1}$

  5. $f(x)=\frac{x-2}{x+2}$

  6. $f(x)=2^x$

  7. $f(x)=3^x$

  1. $f(−2) = 4-2(−2) = 4+4 = 8,$ $f(−1) = 6,$ $f(0) = 4,$ $f(1) = 4-2(1) = 4-2 = 2,$ $f(2) = 0$, $f(a+h) = 4-2a-2h$, $f(a+h)-f(a) = -2h$

  1. $f(−2) = 8(−2)^{2}− 7(−2)+ 3$ $= 8(4)+14+3$ $= 32+14+3$ $= 49,$ $f(−1) = 18,$ $f(0) = 3,$ $f(1) = 8(1)^{2}− 7 (1)+ 3$ $= 8-7+3 = 4,$ $f(2) = 21$, $f(a+h) = 8a^{2}+16ah+8h^{2}-7a-7h+3$, $f(a+h)-f(a)=$ $16ah+8h^{2}-7h$

  1. $f(−2) = -(−2)^{3}+2(−2)$ $= -(−8)-4 $$= 8-4 $$= 4,$ $f(−1)$$ = -(−1)^{3}+2(−1)$$ = -(−1)-2 = -1,$ $f(0) = 0,$ $f(1) = -(1)^{3}+2(1) = 1,$ $f(2) = -4$, $f(a+h) = -a^{3}-3a^{2}h-3ah^{2}+2a-h^{3}+2h$, $f(a+h)-f(a) = -3a^{2}h-3ah^{2}+2h$

  1. $f( - 2) = 3 + \sqrt{( - 2) + 3}$$ = 3 + \sqrt{1}$$ = 3 + 1$$ = 4,$ $\ f( - 1) = \sqrt{2} + 3 \approx 4.41 $,$ f(0) = \sqrt{3} + 3 \approx 4.73,$ $f(1) = 3 + \sqrt{(1) + 3}$$ = 3 + \sqrt{4}$$ = 3 + 2 = 5,$ $f(2) = \sqrt{5} + 3 \approx 5.23$, $f(a + h) = 3 + \sqrt{a + h + 3}$, $f(a + h) - f(a) = \sqrt{a + h + 3} - \sqrt{a + 3}$

  1. $f(−2) = ((−2)−2)((−2)+3)$$ = (−4)(1) = -4,$ $f(−1) = -6$, $f(0) = -6,$ $f(1) = ((1)−2)((1)+3)$$ = (−1)(4) = -4,$ $f(2) = 0$, $f(a+h) = a^{2}+2ah+a+h^{2}+h-6,$ $f(a+h)-f(a) = 2ah+h^{2}+h$

  1. $f( - 2) = \frac{( - 2) - 3}{( - 2) + 1}$$ = \frac{- 5}{- 1} = 5,$ $f( - 1) = \text{undefined},$ $f(0) = - 3,$ $f(1) = - 1,$ $f(2) = - \frac{1}{3},$ $f(a + h) = \frac{a + h - 3}{a + h + 1},\ f(a + h) - f(a) = \frac{4h}{(a + 1)(a + h + 1)}$

  1. $f( - 2) = 2^{- 2} = \frac{1}{2^{2}} = \frac{1}{4},$ $f( - 1) = \frac{1}{2},$ $f(0) = 1,$ $f(1) = 2,$ $f(2) = 4,$ $f(a + h) = 2^{a + h}$, $f(a+h)-f(a) = 2^{a+h}-2^{a}$


  1. Suppose $f(x)=x^2+8x-4$. Compute the following:

    1. $f(−1)+f(1)$

    2. $f(−1)-f(1)$

    1. Using $f(x) = x^{2}+8x-4$: $f(−1) = (−1)^{2}+8(−1)-4 = 1-8-4 = -11$; $f(1) = 1^{2}+8(1)-4 = 1+8-4 = 5$.

    (a) $f(−1)+f(1) = -11+5 = -6$ (b) $f(−1)-f(1) = -11-5 = -16$


  2. Suppose $f(x)=x^2+x+3$. Compute the following:

    1. $f(−2)+f(4)$

    2. $f(−2)-f(4)$

  3. Let $f(t)=3t+5$

    1. Evaluate $f(0)$

    2. Solve $f(t)=0$

    1. Using $f(t) = 3t+5$:

    (a) $f(0) = 3(0)+5 = 5$ (b) $3t+5 = 0$
    $$t = - \frac{5}{3}$$


  4. Let $g(p)=6-2p$

    1. Evaluate $g(0)$

    2. Solve $g(p)=0$

Sketch a graph for each function by creating a table of values.

  1. $f(x)=4x+1$
  2. $g(x)=x^2 -1$
  3. $h(x)=\sqrt{x}+2$
  4. $k(x)=2(x-1)$
    1. $x$$f(x)$
      $-2$$-7$
      $-1$$-3$
      $0$$1$
      $1$$5$
      A straight line through the points (-2, -7), (-1, -3), (0, 1), and (1, 5)
    1. $x$$h(x)$
      $0$$2$
      $1$$3$
      $4$$4$
      A curved line through the points (0,2), (1, 3), and (4,4), continuing up and to the right with an arrow.

  5. Sketch a reasonable graph for each of the following functions. [UW]

    1. Height of a person depending on age.

    2. Height of the top of your head as you jump on a pogo stick for 5 seconds.

    3. The amount of postage you must put on a first class letter, depending on the weight of the letter.

      1. At the beginning, as age increases, height increases. At some point, height stops increasing (as a person stops growing) and height stays the same as age increases. Then, when a person has aged, their height decreases slightly.
      2. As time elapses, the height of a person’s head while jumping on a pogo stick as observed from a fixed point will go up and down in a periodic manner.
      3. The graph does not pass through the origin because you cannot mail a letter with zero postage or a letter with zero weight. The graph begins at the minimum postage and weight, and as the weight increases, the postage increases.

  6. Sketch a reasonable graph for each of the following functions. [UW]

    1. Distance of your big toe from the ground as you ride your bike for 10 seconds.

    2. Your height above the water level in a swimming pool after you dive off the high board.

    3. The percentage of dates and names you’ll remember for a history test, depending on the time you study.

  7. Using the graph shown,

    A curved line connecting the following points: on the bottom left, point K which lines up with a on the x-axis and p on the y-axis. The line touches the x-axis at an x value of b, and the y-axis and a y value of r. Through point L which lines up with c on the x axis and t on the y axis

    1. Evaluate $f$($c$)

    2. Solve $f(x)=p$

    3. Suppose $f(b)=z$. Find $f(z)$

    4. What are the coordinates of points $L$ and $K$?

    1. (a) $   t$ (b) $   x = a$
      (c) $   f(b) = 0$ so $z = 0$. Then $f(z) = f(0) = r$.
      (d) $L = (c,t), K = (a,p)$


  8. Dave leaves his office in Padelford Hall on his way to teach in Gould Hall. Below are several different scenarios. In each case, sketch a plausible (reasonable) graph of the function $s$ = $d$($t$) which keeps track of Dave’s distance $s$ from Padelford Hall at time $t$. Take distance units to be “feet” and time units to be “minutes.” Assume Dave’s path to Gould Hall is long a straight line which is 2400 feet long. [UW]

    A box on the left labeled padelford. A stick figure is just to the right of padelford, with an arrow pointing right. on the far right, a box labeled gould.

    1. Dave leaves Padelford Hall and walks at a constant spend until he reaches Gould Hall 10 minutes later.

    2. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute. He then continues on to Gould Hall at the same constant speed he had when he originally left Padelford Hall.

    3. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave then continues on to Gould Hall at twice the constant speed he had when he originally left Padelford Hall.

    4. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave is totally lost, so he simply heads back to his office, walking the same constant speed he had when he originally left Padelford Hall.

    5. Dave leaves Padelford heading for Gould Hall at the same instant Angela leaves Gould Hall heading for Padelford Hall. Both walk at a constant speed, but Angela walks twice as fast as Dave. Indicate a plot of “distance from Padelford” vs. “time” for the both Angela and Dave.

    6. Suppose you want to sketch the graph of a new function s=g(t) that keeps track of Dave’s distance s from Gould Hall at time t. How would your graphs change in (a)-(e)?