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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.)

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.

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.

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 \( \{ x ~|~ x \geq 0 \} \).

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.

Practice

Unless otherwise instructed, find the domain and the range from the graphs.

• Practice 427
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

PatrickJMT - 427 video solution

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• Practice 428
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

PatrickJMT - 428 video solution

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• Practice 429
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

PatrickJMT - 429 video solution

video by PatrickJMT

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• Practice 430
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

PatrickJMT - 430 video solution

video by PatrickJMT

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• Practice 432
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

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 and those values do not affect the problem solution.

Krista King Math - 432 video solution

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• Practice 1308
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

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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• Practice 431
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

PatrickJMT - 431 video solution

video by PatrickJMT

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• Practice 1309
• Solution

Problem Statement

Find the domain and the range from the graph.

Solution

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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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.

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. Here are a few examples to emphasize this.

- 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.

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.

Practice

Unless otherwise instructed, find the domain and the range of these functions.

• Practice 433
• Solution

\(\displaystyle{f(x)=\frac{5}{x-2}}\)

Problem Statement

Find the domain of \(\displaystyle{f(x)=\frac{5}{x-2}}\)

Solution

The Organic Chemistry Tutor - 433 video solution

video by The Organic Chemistry Tutor

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• Practice 434
• Solution

\(\displaystyle{f(x)=\frac{1}{x^2-x-6}}\)

Problem Statement

Find the domain of \(\displaystyle{f(x)=\frac{1}{x^2-x-6}}\)

Solution

434 video solution

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• Practice 3864
• Solution

\(\displaystyle{ f(x) = \frac{3x-8}{x^2-9x+20} }\)

Problem Statement

Find the domain of \(\displaystyle{ f(x) = \frac{3x-8}{x^2-9x+20} }\)

Solution

The Organic Chemistry Tutor - 3864 video solution

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• Practice 3865
• Solution

\(\displaystyle{ f(x) = \frac{2x-3}{x^2+4} }\)

Problem Statement

Find the domain of \(\displaystyle{ f(x) = \frac{2x-3}{x^2+4} }\)

Solution

The Organic Chemistry Tutor - 3865 video solution

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• Practice 445
• Solution

\( f(x) = \sqrt{x-4} \)

Problem Statement

Find the domain of \( f(x) = \sqrt{x-4} \)

Solution

The Organic Chemistry Tutor - 445 video solution

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• Practice 3866
• Solution

\( f(x) = \sqrt{x^2+3x-28} \)

Problem Statement

Find the domain of \( f(x) = \sqrt{x^2+3x-28} \)

Solution

The Organic Chemistry Tutor - 3866 video solution

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• Practice 435
• Solution

\(\displaystyle{f(x)=\frac{\sqrt{x-1}}{x^2+4}}\)

Problem Statement

Find the domain of \(\displaystyle{f(x)=\frac{\sqrt{x-1}}{x^2+4}}\)

Solution

435 video solution

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• Practice 437
• Solution

\(f(x)=\ln(x-8)\)

Problem Statement

Find the domain of \(f(x)=\ln(x-8)\)

Solution

437 video solution

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• Practice 1908
• Answer
• Solution

\(f(x)=10^{2x}+\ln(21-3x)\)

Problem Statement

Find the domain and range of \(f(x)=10^{2x}+\ln(21-3x)\)

Final Answer

domain: \(x < 7\); range: \(y > 3.17771\)

Problem Statement

Find the domain and range of \(f(x)=10^{2x}+\ln(21-3x)\)

Solution

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).

Krista King Math - 1908 video solution

video by Krista King Math

Final Answer

domain: \(x < 7\); range: \(y > 3.17771\)

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• Practice 443
• Solution

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

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

Krista King Math - 443 video solution

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• Practice 436
• Solution

\(\displaystyle{f(x)=\frac{1}{\sqrt{x^2-4}}}\)

Problem Statement

Find the domain of \(\displaystyle{f(x)=\frac{1}{\sqrt{x^2-4}}}\)

Solution

436 video solution

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• Practice 3867
• Solution

\(\displaystyle{ f(x) = \frac{2x-7}{\sqrt{x+3}} }\)

Problem Statement

Find the domain of \(\displaystyle{ f(x) = \frac{2x-7}{\sqrt{x+3}} }\)

Solution

The Organic Chemistry Tutor - 3867 video solution

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• Practice 3868
• Solution

\(\displaystyle{ f(x) = \frac{\sqrt{x-4}}{x^2-25} }\)

Problem Statement

Find the domain of \(\displaystyle{ f(x) = \frac{\sqrt{x-4}}{x^2-25} }\)

Solution

The Organic Chemistry Tutor - 3868 video solution

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• Practice 3869
• Solution

\(\displaystyle{ f(x) = \frac{\sqrt{x+3}}{\sqrt{x^2-16}} }\)

Problem Statement

Find the domain of \(\displaystyle{ f(x) = \frac{\sqrt{x+3}}{\sqrt{x^2-16}} }\)

Solution

The Organic Chemistry Tutor - 3869 video solution

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