This section covers:
 Algebraic Functions Versus Relations
 Vertical Line Test
 Domain and Range of Relations and Functions
 Finding the Domain Algebraically
 More Practice
Note: More advanced topics with functions can be found in the Parent Functions and Transformations and Advanced Functions: Compositions, Even and Odd, and Extrema section. End behavior of functions can be found here in the Parent Functions and Transformations section, here in the Graphing and Finding Roots of Polynomial Functions section, and also in the Graphing Rational Functions, including Asymptotes section. Intermediate Value Theorem (IVT) can be found here in the Limits and Continuity section.
Algebraic Functions Versus Relations
When we first talked about the coordinate system, we worked with the graph that shows the relationship between how many hours we worked (the independent variable, or the “\(x\)”), and how much money we made (the dependent variable, or the “\(y\)”). Any relationship between two variables, where one depends on the other, is called a relation, since it relates two things.
This particular relation is an algebraic function, since there is only one \(y\) for each \(x\). In other words, since the \(x\) is the “question” and \(y\) is the “answer”, we can only have one answer for each question. For whatever is the number of hours we work, we only get paid a certain amount for that: \(y=10x\).
Again, a function is just a fancy way of saying something depends on something else, and there’s only one “\(y\)” for every “\(x\)”. But the other thing you’ll learn about functions is that they can be written a funny way; a way that looks really complicated, but they are just trying to confuse us – it’s not that bad!
Instead of our original equation, \(y=10x\), we can write it like this: \(f(x)=10x\).
Note that this is not “\(f\) times \(x\)”; it is “\(f\) of \(x\)”. What it means is that \(x\) is on the right hand sign of the “=” sign, and you can put different values in for \(x\) on the left hand side to get one and only one value on the righthand sign. So again, “\(f(x)\)” is really “\(y\)”. It’s that simple.
Here are some examples of plugging in things on the left hand side, and then, to get our answer, we plug what that is for every \(x\)on the righthand side:
\(\begin{align}f\left( x \right)&=10x\text{ (original function)}\\f\left( 2 \right)&=10\times 2=20\\f\left( 9 \right)&=10\times 9=90\\f\left( t \right)&=10t\\f\left( {x+1} \right)&=10\left( {x+1} \right)=10x+10\end{align}\)  Just remember that the lefthand side is always \(f\left( {\text{something}} \right)\), and just plug that something into every instance of the variable on the right – every place there is an \(x\).
In this example, we only have one place to plug in for the \(x\). 
Again, what makes a relation a function is that you can only have one “answer” (the \(y\)) for each “question” (the \(x\)). All functions are relations, but not all relations are functions.
Vertical Line Test
Notice that when we have a function, we can’t draw a vertical that goes through more than one point. This is called the vertical line test, and it’s a useful tool to determine if a graph is a function or not. For example, we can tell the following graph is not a function since we can draw a vertical line and hit more than one point:
Graphs – NOT Functions  Notes 
See how we have two “answers” for \(x=4\)?
The points \(\left\{ {\left( {2,1} \right),\left( {3,0} \right),\left( {4,2} \right),\left( {4,2} \right)} \right\}\) are not a function.
We can draw a vertical line and see that we have two points on the same line: not a function.
Notice that it’s not a function when we have 2 different \(y\) values for the same \(x\) value. 

See how when \(x=4\) (the “question”), we have two points: \((4,2)\), and \((4,–2)\). Thus, we have two “answers”.
This would mess up things in real life (like in the engineering or banking fields), since functions in real life must give us only one answer when we ask a question (like how much we have in the bank on a certain day). 
Here are some examples of functions. Notice that you can have two “questions” (\(x\)) for the same “answer” (\(y\)):
Graphs – Functions  Notes 
See how we have two “questions”, or 2 \(x\) values with the same “answer” (\(y\) value)? \(x\) values 2 and 4 have the same \(y\) value that is 1.
And see how if we were to draw any vertical line, we’d never cross two points?
The points \(\left\{ {\left( {1,2} \right)\left( {2,1} \right),\left( {3,0} \right),\left( {4,1} \right)} \right\}\) are a function.
Like the first graph of points above, the function is called discrete, since you have to pick up your pencil to get from one point to another. Otherwise, functions are continuous – if you never have to pick up your pencil for the whole function. 

See how if we were to draw any vertical line, we’d never cross two points? And see how we’d never need to pick up our pencil when drawing this graph?
Try it yourself!
This relation is a continuous function. 
Another way to look at a set of points and determine whether or not they are functions is to draw what we call mapping diagrams, since we are mapping the \(x\) values to the \(y\) values. We order values from smallest to largest and don’t repeat the values on each side and match them up. If we have more than one \(y\) value for one \(x\) value, we don’t have a function. Here are some examples:
Points  Mapping Diagram  Function? Yes or No 
\(\left\{ {\left( {2,1} \right),\left( {3,0} \right),\left( {4,2} \right),\left( {4,2} \right)} \right\}\)  Not a function: we have 2 \(y\)’s (–2 and 2) for the same \(x\) (4).
It’s not OK to have one question with 2 different answers. 

\(\left\{ {\left( {1,2} \right)\left( {2,1} \right),\left( {3,0} \right),\left( {4,1} \right)} \right\}\)  Is a function: we have 2 \(x\)’s for the same \(y\), but this is fine.
It’s OK to have 2 questions with the same answer. 
Domain and Range of Relations and Functions
Domain and Range of functions (and relations) sound really difficult and scary, but they are not really bad at all. You know how those mathematicians like to use fancy words for easy stuff?
Remember that since “d” comes before “r”, the domain of functions have to do with the “\(x\)”’s and the range of functions have to do with the “\(y\)”’s. To get the domain, we are just looking for all the possible values of \(x\) for that function (from smallest to largest), and for the range, we are looking for all possible values of \(y\) for that function (again, from smallest to largest).
To help me do this, I like to use my pencil – but it’s backwards compared to what you might think. To find the domain, I put my pencil vertically and start at the left and see where it first hits a point. Then I push it through all the way to the right to see where it ends hitting points.
For the range, I do the same thing, but with a horizontal pencil that’s moving up:
Graphs  Notes 
When we move the vertical pencil from left to right, we get the following \(x\) values: 1, 2, 3, and 4.
The domain is \(\left\{ {1,2,3,4} \right\}\). Note that since we are only dealing with points in this graph, we write the set of \(x\) values in a set with the brackets.
Similarly, when we move the horizontal pencil from the bottom to the top, we get the following \(y\) values: –2, 0, and 1.
The range is \(\left\{ {2,0,1} \right\}\). Note that we don’t need to repeat values. 
Here are more examples, using what we call “Interval Notation”. (We saw this in the Inequalities Section). This is the most commonly used way to describe domains and ranges, and it always goes from lowest to highest with “(“ (soft brackets) if the relation or function doesn’t hit the point, and “[“ (hard brackets) if the relation or function does hit the line. If you have to skip over any numbers, you do so by using the “U” sign, which means union, or putting things together.
We can also use Set Builder/Inequality Notation, where, as we saw before, we use inequality signs to describe the ranges.
Notice on when we see arrows in the graphs, we have to assume that the function “goes on forever” in those directions.
Function  Domain/Range 
Interval Notation: Domain: \(\left( {\,\infty ,\infty } \right)\) (or all Real Numbers, \(\mathbb{R}\)) Range: \(\left[ {4,\infty } \right)\)
Inequality Notation: Domain: \(\,\infty <x<\infty \,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\left\{ {x:x\in \mathbb{R}} \right\}\) Range: \(\left\{ {y\,y\ge 4} \right\}\)
Notice that the graph goes through \((0,–4)\), so –4 is included in the range. 

Interval Notation: Domain: \(\left[ {9,1} \right)\cup \left( {1,\infty } \right)\) Range: \(\displaystyle \left( {\infty ,3} \right)\cup \left[ {0,5} \right)\)
Inequality Notation: Domain: \(\left\{ {x\,9\le x<1\,\,\text{or}\,\,x>1} \right\}\) Range: \(\left\{ {y\,\,y<3\,\,\,\text{or}\,\,\,0\le y<5} \right\}\)
Note that the arrow indicates that the domain and range go on forever in the rightdown direction. Also note that since the \(x\)’s don’t include “1”, we have to “jump over” it using a union (\(\cup \)) sign. We also have to use the union sign in a range. 

Interval Notation: Domain: \(\left[ {6,1} \right)\cup \left[ {0,5} \right)\,\) Range: \(\left( {\infty ,7} \right]\)
Inequality Notation: Domain:\(\left\{ {x\,6\le x<1\,\,\,\text{or}\,\,\,0\le x<5} \right\}\) Range: \(\left\{ {y\,\,\,y\le 7} \right\}\)
Note that the domain skips from 1 to 0 (but includes 0). It gets closer and closer to 5 (an asymptote), but never reaches it. The range goes all the way up to 7, since there is overlap between parts of the function.

If you don’t see how we got the domain and range above, use the pencil trick, and make sure you start from the left for the domain (with vertical pencil) and from the bottom with the range (with horizontal pencil).
Note again that the last two graphs are not functions; they do not meet the Vertical Line Test requirements.
Restricted Domains: Finding the Domain Algebraically
In many cases, the domain is restricted:
 It is randomly indicated that way in the problem. For example, \(f\left( x \right)=3x1,\,\,x\ge 0\).
 There is a variable in the denominator and that denominator could be 0. For example, \(\displaystyle f\left( x \right)=\frac{1}{{x3}}\). (In this case, \(x3\ne 0;\,\,\,\,\,x\ne 3\)).
 There is a variable underneath an even radical sign, and that radicand (underneath the radical sign) could be negative. For example, \(f\left( x \right)=\sqrt{{x+4}}\). (In this case, \(x+4\ge 0;\,\,\,\,\,\,x\ge 4\)).
 (More advanced – see Logarithmic Functions section) If there’s a variable in the argument of a log or ln function; log arguments must be greater than 0. For example, \(f\left( x \right)=\log \left( {8x} \right)\). (In this case, \(8x>0;\,\,\,\,x<8\)).
(There are other types of functions, like trigonometric functions, that have domain restrictions, but we won’t address these here.)
(Note that if we could have a mixture of the above restrictions, for example, for \(\displaystyle f\left( x \right)=\frac{{\sqrt{x}}}{{x1}}\), \(x\ge 0\) and \(x\ne 1\), so the domain of \(x\) is \(\left[ {0,1} \right)\cup \left( {1,\infty } \right)\).)
We start out assuming that the domain of a function is all real numbers, but then see if there are any exceptions, as seen in the table. We will learn more about rational functions (shown in the first two examples, where there are variables in the denominator) in the Rational Functions and Equations, and Graphing Rational Functions, including Asymptotes sections.
Exceptions to All Reals in Domain  How to Get the Restricted Domain 
Is there anywhere in the function where there is an \(\boldsymbol {x}\) in a denominator, and that denominator could somehow be 0?
We haven’t learned these types of functions yet, where we can have a variable in the denominator, but we will in the Rational Functions, Equations and Inequalities section. 
If so, the domain is all real numbers, excluding where any denominator could be 0; this is because we can never divide by 0.
Example:\(\displaystyle \color{#800000}{{f(x)=\frac{x}{{x3}}}}\text{: }\,\,x3\ne 0\text{, therefore }x\ne 3\)
We solve the equation where \(x3=0\)’ whatever we get for \(x\) can’t be in the domain. Therefore, the domain is all real numbers except for 3, or \(\left( {\infty ,3} \right)\cup \left( {3,\infty } \right)\).
Note that if we had \(\displaystyle f(x)=\frac{x}{{{{x}^{2}}+1}}\) for example, this wouldn’t have a restriction; since the bottom could never be negative (try to make it negative!). 
Is there anywhere in the function where \(\boldsymbol {x}\) is inside an even radical sign (root function?), and that radicand could somehow be negative?
If we remember from the Powers, Exponents, Radicals, and Scientific Notation section, we can’t take an even root of a negative number, since we can’t multiply something an even number of times and get a negative number. 
If so, set what is under any even radical sign to be \(\ge 0\), since we can’t take the even root of a negative number. The domain is all real numbers, excluding where \(x\) could make something under an even radical sign negative.
Example: \(\displaystyle \color{#800000}{{f(x)=\frac{x}{{\sqrt{{x3}}}}}}\text{: }\,\,x3\ge 0\text{, but also }\sqrt{{x3}}\ne 0\text{ }\)
We had to use both exceptions, since we have both an even root and also a denominator with an \(x\) in it. Solving both equations, we find that \(x\) has to be greater than or equal to 3, but also cannot be 3. Therefore, the domain is all real numbers greater than 3, or \(\left( {3,\infty } \right)\).
Note that if we had \(f(x)=\sqrt{{{{x}^{2}}+1}}\) for example, this wouldn’t have a restriction; since what’s under the radical sign (the radicand) can’t be negative (try to make it negative!). 
Does it indicate anywhere in the problem that the domain is restricted?  Example: \(f(x)=5x,\,\,\,\,\,x\ge 0\)
I know this seems obvious, but in many cases (especially in real world problems), the domain is restricted, even if there is no variable in a denominator or under an even root sign.
This domain is all real numbers greater than 0, or \(\left( {0,\infty } \right)\). In case, the range would be all real numbers greater than 5, or \(\left( {5,\infty } \right)\) (plug in the domain to see this). 
Is there a variable in the argument of a log (or ln) function? Could the argument ever be negative?
We haven’t learned these types of functions yet, but we will in the Logarithmic Functions section. 
Example: \(\color{#800000}{{f(x)=\ln \left( {x+4} \right)}}:\,\,\,\,\,x+4>0\text{, therefore }x>4\)
Log arguments must be greater than 0. We solve the equation where \(x+4>0\); whatever we get for \(x\) can’t be in the domain.
Therefore, the domain is all real numbers greater than –4, or \(\displaystyle \left( {4,\infty } \right)\).

We will work on more advanced topics with functions later, in the Advanced Functions section.
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