17Calculus Precalculus - Domain and Range
The domain is the set of all allowable values that a function can accept as input and produce a meaningful value.
The range is the set of all meaningful values that come out of a function.
Note - Discussion on the domain of composite functions can be found on the composite functions page.
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Why Domain Is Important In Calculus
A function cannot be defined without a domain. But wait! Teachers (and textbooks) do it all the time, right?! Well . . . it may seem like it, but that's not really what is going on. Most teachers (and textbooks) don't come right out and say what the domain is. However, if they don't, that just means that the domain is implied by the equation. (One of the most common things you will see in math textbooks and classes is not stating what the authors and teachers think is obvious that may not be obvious to you. One of your jobs while learning math is to fill in the blanks.)
Here is why domain is important in calculus.
Everything in calculus requires the understanding of the domain since the domain (and sometimes the range too) will often determine if and when you can apply a theorem or use a technique. So to be able to do calculus problems, you absolutely need to know the domain. Every fully-defined function includes domain, whether the domain is stated or not.
Most of the time the domain is implied and not explicitly stated. However, don't let that fool you. Important: Two equations with different domains are considered different functions, even if one or both domains are not explictly stated. Keep this in mind when you are working with limits. Good teachers will emphasize this and require your notation to reflect this knowledge. Let's look at an example.
Example
Assuming we are working with only real numbers (this is true most of the time in calculus, unless otherwise stated or implied by the context), let's compare these two functions.
Function |
Domain | |
|---|---|---|
\(\displaystyle{ f(x) = \frac{(x+1)(x-1)}{(x+1)} }\) |
\( \{ x | x \in \mathbb{R} ~ and ~ x \neq -1 \} \) | |
\(\displaystyle{ g(x) = x-1 }\) |
\( \{ x | x \in \mathbb{R} \} \) |
If you look closely, you can see that you can cancel \(x+1\) in \(f(x)\) to get \(g(x)\). However, \(f(x) \neq g(x)\). Why? Because the domains are different. The domain of \( f(x) \) is all real numbers except for \(x = -1\) The domain of \(g(x)\) is all real numbers. In calculus, you can never just say \( f(x) = g(x)\) for these two functions.
However, you CAN say that \( f(x) = g(x) \) IF you also say \(x \neq -1\), but this must be explicitly stated since it is not obvious for \(g(x)\). So be very careful.
Limits
Another thing you can say about these two functions is
\[\displaystyle{ \lim_{x \rightarrow -1}{f(x)} = \lim_{x \rightarrow -1}{g(x)} }\]
and you don't need to say anything about what is going on at \(x = -1\). This is true because of the limit key explained on the 17calculus limits page.
However, in this equation we are talking about the LIMITS being equal, not the functions themselves being equal, an important distinction.
What Are The Domain and Range?
To get started, we need to understand what the domain and range are. Most students think they know but take a few minutes to watch this video to make sure you really DO know. Then you can move forward with more confidence.
Dr Chris Tisdell - Domain and range of a function [13min-59secs]
video by Dr Chris Tisdell |
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For example, if the function is \( f(x) = \ln(x) \) and we want only real numbers to come out, we can't put in zero or any negative number for \(x\). Therefore, the domain is \( \{ x ~|~ x > 0 \} \). Another example is the square root function. We are allowed to take square roots of positive numbers and zero but, for calculus (since we usually work only with real numbers), we do not take square roots of negative numbers. So the domain of the square root function is \( \{ x ~|~ x \geq 0 \} \).
Finding Domain and Range From A Graph
This is the easiest case since a graph will tell us a lot about a function. For the domain, we just scan left-to-right and see if the graph is above, below or touching the x-axis, in which case, the x-values are in the domain. Similarly for the range, we scan vertically to see if the graph is to the left, right or touching the y-axis and, if so, then the y-values are in the range. Here are some practice problems to help you understand this better.
Practice
Unless otherwise instructed, find the domain and the range from the graphs.
Problem Statement |
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Find the domain and the range from the graph.
Solution |
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Problem Statement |
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Solution |
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The graph given in the problem statement is slightly different than she has in the video. However, the domain and range are the same in both cases. Only the values of f(1) and f(5) differ. But you were not asked to find those in the problem statement.
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Problem Statement |
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Find the domain and the range from the graph.
Solution |
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From the graph, you can see that there are asymptotes at \(x=-1\) and \(x=2\). All other x-values will work. So the domain is all real numbers except for \(-1\) and \(2\). The range is all real numbers.
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Problem Statement |
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Find the domain and the range from the graph.
Solution |
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Problem Statement |
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Find the domain and the range from the graph.
Solution |
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From the graph, you can see that there are vertical asymptotes at \(x=-1\) and \(x=2\). Also notice there are no x-values in the interval \(-1 < x < 2\). So the domain is \(\{ x ~|~ x < -1 \cup x > 2 \}\).
The range is a bit more challenging. At the vertical asymptotes, the graph goes off to infinity in both the positive and negative direction. However, around \(y=1\), it is not clear what is going on. You should have enough experience with horizontal asymptotes to be able to recognize that there is a horizontal asymptote at \(y=1\). So the range is all real numbers except for \(1\).
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Finding Domain and Range From The Equation
Let's start by discussing how to determine the domain of a function from the equation. (Finding it from the graph is easier but in calculus you need to be able to work only from the equation.) You need to know the domain before you can find the range. The domain is a list of \(x\)-values that are allowed to be put into the function. As you should already know, some functions are just not defined for certain values. When you try to put those values into a function, either nothing comes out or what does, doesn't make sense.
Steps To Determine Domain
Rather than trying to figure out what WILL work, it is much easier to determine what you can't put into a function. So, to determine the domain, we start with the set of all real numbers (in calculus, we usually stay with real numbers unless explicitly stated otherwise) and then remove the numbers that don't work. What is left is the domain. Here are the cases you will come across most of the time of situations that you don't want to have as input values.
1. In rational expressions, you get zero in the denominator.
2. In even roots, you have negative numbers (zero is okay).
3. In logarithms, you have numbers less than or equal to zero.
4. In the inverse trig functions arcsine and arccosine, values greater than 1 or less than -1.
The idea is to start out with all real numbers and remove numbers that fall under one of the above cases. What remains is the domain. Here are a few important things to remember.
- The domain of polynomials is the set of all real numbers.
- When combining terms, find the domain of each piece and then take the intersection of the sets to get the domain of the entire expression.
Here are a few functions and their domains. Knowing these should be enough for you to work most calculus problems.
function type |
domain | |
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polynomials |
all real numbers | |
rational functions |
all real numbers except where the denominator is zero | |
root functions |
even roots: real numbers greater than or equal to zero; | |
trig functions |
sine and cosine: all real numbers | |
exponential functions |
all real numbers | |
logarithm functions |
real numbers greater than zero (not including zero) |
Finding The Range
Once you have the domain, you can find the range. The range is not as easily determined. I usually use a combination of the domain, the graph and these simple rules.
1. Even roots are always positive.
2. Logarithms can have both positive and negative output.
Okay, try out these ideas on these practice problems.
Practice
Unless otherwise instructed, find the domain and the range of these functions.
\(\displaystyle{f(x)=\frac{1}{x-2}}\)
Problem Statement |
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\(\displaystyle{f(x)=\frac{1}{x-2}}\)
Solution |
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\(\displaystyle{f(x)=\frac{1}{x^2-x-6}}\)
Problem Statement |
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\(\displaystyle{f(x)=\frac{1}{x^2-x-6}}\)
Solution |
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\(f(x)= \sqrt{2x-8}\)
Problem Statement |
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\(f(x)= \sqrt{2x-8}\)
Solution |
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\(\displaystyle{f(x)=\frac{\sqrt{x-1}}{x^2+4}}\)
Problem Statement |
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\(\displaystyle{f(x)=\frac{\sqrt{x-1}}{x^2+4}}\)
Solution |
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\(\displaystyle{f(x)=\ln(x-8)}\)
Problem Statement |
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\(\displaystyle{f(x)=\ln(x-8)}\)
Solution |
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\(f(x)=10^{2x}+\ln(21-3x)\)
Problem Statement |
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Find the domain and range of \(f(x)=10^{2x}+\ln(21-3x)\)
Final Answer |
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domain: \(x < 7\); range: \(y > 3.17771\)
Problem Statement |
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Find the domain and range of \(f(x)=10^{2x}+\ln(21-3x)\)
Solution |
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For the range in this problem, the presenter states that the range is \(y > 0\). However, if you graph the function, as we have done below, you can see that the range is actually closer to \(y > 3\). With the training you have in precalculus, this is pretty much as close as you can get using equations and looking at the graph. However, if you graph the function and then have the calculator find the minimum, it occurs at \(x \approx -0.77703\) giving the range closer to \(y > 3.17771\) (approximately).
1908 video
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Final Answer |
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domain: \(x < 7\); range: \(y > 3.17771\) |
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For \(f(x)=x^3+2x^2\) and \(g(x)=3x^2-1\), find the domains of \((f+g)(x),\) \((f-g)(x), (f\cdot g)(x), (f/g)(x)\).
Problem Statement |
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For \(f(x)=x^3+2x^2\) and \(g(x)=3x^2-1\), find the domains of \((f+g)(x),\) \((f-g)(x), (f\cdot g)(x), (f/g)(x)\).
Solution |
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\(\displaystyle{f(x)=\frac{1}{\sqrt{x^2-4}}}\)
Problem Statement |
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\(\displaystyle{f(x)=\frac{1}{\sqrt{x^2-4}}}\)
Solution |
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Really UNDERSTAND Precalculus
Related Topics and Links
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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