What are Functions? Excerpt from “Math Dialogue: Functions”


What are functions? I could just insert the standard definition here, but I fear that this might not be the best approach for those who have just started their journey into the fascinating world of mathematics. For one, any common textbook will include the definition, so if that’s all you’re looking for, you don’t need to continue reading here. Secondly, it is much more rewarding to build towards the definition step by step. This approach minimizes the risk of developing deficits and falling prey to misunderstandings.


So where do we start?


We have two options here. We could take the intuitive, concept-focused approach or the more abstract, mathematically rigorous path. My recommendation is to go down both roads, starting with the more intuitive approach and taking care of the strict details later on. This will allow you to get familiar with the concept of the function and apply it to solve real-world problems without first delving into sets, Cartesian products as well as relations and their properties.


Sounds reasonable.


Then let’s get started. For now we will think of a function as a mathematical expression that allows us to insert the value of one quantity x and spits out the value of another quantity y. So it’s basically an input-output system.


Can you give me an example of this?


Certainly. Here is a function: y = 2·x + 4. As you can see, there are two variables in there, the so-called independent variable x and the dependent variable y. The variable x is called independent because we are free to choose any value for it. Once a value is chosen, we do what the mathematical expression tells us to do, in this case multiply two by the value we have chosen for x and add four to that. The result of this is the corresponding value of the dependent variable y.


So I can choose any value for x?


That’s right, try out any value.


Okay, I’ll set x = 1 then. When I insert this into the expression I get: y = 2·1 + 4 = 6. What does this tell me?


This calculation tells you that the function y = 2·x + 4 links the input x = 1 with the output y = 6. Go on, try out another input.


Okay, can I use x = 0?


Certainly. Any real number works here.


For x = 0 I get y = 2·0 + 4 = 4. So the function y = 2·x + 4 links the input x = 0 with the output y = 4.


That’s right. Now it should be clear why x is called the independent variable and y the dependent variable. While you may choose any real number for x, sometimes there are common sense restrictions though, we’ll get to that later, the value of y is determined by the form of the function. A few more words on terminology and notation. Sometimes the output is also called the value of the function. We’ve just found that the function y = 2·x + 4 links the input x = 1 with the output y = 6. We could restate that as follows: at x = 1 the function takes on the value y = 6. The other input-output pair we found was x = 0 and y = 4. In other words: at x = 0 the value of the function is y = 4. Keep that in mind.

As for notation, it is very common to use f(x) instead of y. This emphasizes that the expression we are dealing with should be interpreted as a function of the independent variable x. It also allows us to note the input-output pairs in a more compact fashion by including specific values of x in the bracket. Here’s what I mean.

For the function we can write: f(x) = 2·x + 4. Inserting x = 1 we get: f(1) = 2·1 + 4 = 6 or, omitting the calculation, f(1) = 6. The latter is just a very compact way of saying that for x = 1 we get the output y = 6. In a similar manner we can write f(0) = 4 to state that at x = 0 the function takes on the value y = 4. Please insert another value for x using this notation.


Will do. I’ll choose x = -1. Using this value I get: f(-1) = 2·(-1) + 4 = 2 or in short f(-1) = 2. So at x = -1 the value of the function is y = 2. Is all of this correct?


Yes, that’s correct.


You just mentioned that sometimes there are common sense restrictions for the independent variable x. Can I see an example of this?


Okay, let’s get to this right now. Consider the function f(x) = 1/x. Please insert the value x = 1.


For x = 1 I get f(1) = 1/1 = 1. So is it a problem that the output is the same as the input?


Not at all, at x = 1 the function f(x) = 1/x takes on the value y = 1 and this is just fine. The input x = 2 also works well: f(2) = 1/2, so x = 2 is linked with the output y = 1/2. But we will run into problems when trying to insert x = 0.


I see, division by zero. For x = 0 we have f(0) = 1/0 and this expression makes no sense.


That’s right, division by zero is strictly verboten. So whenever an input x would lead to division by zero, we have to rule it out. Let’s state this a bit more elegantly. Every function has a domain. This is just the set of all inputs for which the function produces a real-valued output. For example, the domain of the function f(x) = 2·x + 4 is the set of all real numbers since we can insert any real number x without running into problems. The domain of the function f(x) = 1/x is the set of all real numbers with the number zero excluded since we can use all real numbers as inputs except for zero.

Can you see why the domain of the function f(x) = 1/(3·x – 12) is the set of all real numbers excluding the number four? If it is not obvious, try to insert x = 4.


Okay, for x = 4 I get f(4) = 1/(3·4 – 12) = 1/0. Oh yes, division by zero again.


Correct. That’s why we say that the domain of the function f(x) = 1/(3·x – 12) is the set of all real numbers excluding the number four. Any input x works except for x = 4. So whenever there’s an x somewhere in the denominator, watch out for this. Sometimes we have to exclude inputs for other reasons, too. Consider the function f(x) = sqrt(x). The abbreviation “sqrt” refers to the square root of x. Please compute the value of the function for the inputs x = 0, x = 1 and x = 2.


Will do.

f(0) = sqrt(0) = 0

At x = 0 the value of the function is 0.

f(1) = sqrt(1) = 1

At x = 1 the value of the function is 1.

f(2) = sqrt(2) = 1.4142 …

At x = 2 the value of the function is 1.4142 … All of this looks fine to me. Or is there a problem here?


No problem at all. But now try x = -1.


Okay, f(-1) = sqrt(-1) = … Oh, seems like my calculator spits out an error message here. What’s going on?


Seems like your calculator knows math well. There is no square root of a negative number. Think about it. We say that the square root of the number 4 is 2 because when you multiply 2 by itself you get 4. Note that 4 has another square root and for the same reason. When you multiply -2 by itself, you also get 4, so -2 is also a square root of 4.

Let’s choose another positive number, say 9. The square root of 9 is 3 because when you multiply 3 by itself you get 9. Another square root of 9 is -3 since multiplying -3 by itself leads to 9. So far so good, but what is the square root of -9? Which number can you multiply by itself to produce -9?


Hmmm … 3 doesn’t work since 3 multiplied by itself is 9, -3 also doesn’t work since -3 multiplied by itself is 9. Looks like I can’t think of any number I could multiply by itself to get the result -9.


That’s right, no such real number exists. In other words: there is no real-valued square root of -9. Actually, no negative number has a real-valued square root. That’s why your calculator complained when you gave him the task of finding the square root of -1. For our function f(x) = sqrt(x) all of this means that inserting an x smaller than zero would lead to a nonsensical result. We say that the domain of the function f(x) = sqrt(x) is the set of all positive real numbers including zero.

In summary, when trying to determine the domain of a function, that is, the set of all inputs that lead to a real-valued output, make sure to exclude any values of x that would a) lead to division by zero or b) produce a negative number under a square root sign. Unless faced with a particularly exotic function, the domain of the function is then simply the set of all real numbers excluding values of x that lead to division by zero and those values of x that produce negative numbers under a square root sign.

I promise we will get back to this, but I want to return to the concept of the function before doing some exercises. Let’s go back to the introductory example: f(x) = 2·x + 4. Please make an input-output table for the following inputs: x = -3, -2, -1, 0, 1, 2 and 3.

This was an excerpt from “Math Dialogue: Functions“, available on Amazon for Kindle.

What are Functions? A Short and Simple Explanation

Understanding functions is vital for anyone intending to master calculus or learning mathematical physics. For those who have never heard of the concept of the function, here’s a quick introduction. A function f(x) is a mathematical expression, you can think of it as an input-output system, establishing a connection between one independent variable x and a dependent variable y. We “throw” in a certain value of x, do what the mathematical expression f(x) demands us to do, and get a corresponding value of y in return. Here’s an example of this:

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The expression on the right tells us that when given a certain value of x, we need to multiply the square of x by 2, subtract 3x from the result and in a final step add one. For example, using x = 2, the function returns the value:

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So this particular function links the input x = 2 to the output y = 3. This is called the value of the function f(x) at x = 2. Of course, we are free to choose any other value for x and see what the function does with it. Inserting x = 1, we get:

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So given the input x = 1, the function produces the output y = 0. Whenever this happens, a value of zero is returned, we call the respective value of x a root of the function. So the above function has one root at x = 1. Let’s check a few more values:

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The first line tells us that for x = 0, the value of the function lies on y = 1. For x = -1 we get y = 6 and for x = -2 the result y = 15. What to do with this? For one, we can interpret these values geometrically. We can consider any pair of x and y as a point P(x / y) in the Cartesian coordinate system. Since we could check every x we desire and calculate the corresponding value of y using the function, the function defines a graph in the coordinate system. The graph of the above function f(x) will go through the points P(2 / 3), P(1 / 0), P(0 / 1), P(-1, 6) and P(-2 / 15). Here’s the plot:

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Graph of f(x) = 2x² – 3x + 1

You can confirm that the points indeed lie on the graph by following the x-axis, as usual the horizontal axis, and determining what distance the curve has from the x-axis at a certain value of x. For example, to find the point P(2 / 3), we move, starting from the origin, two units to the right along the x-axis and then three units upwards, in direction of the y-axis. Here we meet the curve, confirming that the graph includes the point P(2 / 3). Make sure to check this for all other points we calculated. Of course, to produce such a neat plot, we need to insert a lot more than just six values for x. This uncreative work is best done by a computer. Feel free to check out the easy-to-use website for this, it doesn’t cost a thing and requires no registration. Note that the plot also shows a second root at x = 0.5, the point P(0.5 / 0). Let’s make sure that this value of x is a root of our function f(x) by inserting x = 0.5 and hoping that it produces the output y = 0:

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As expected. This was all very mathematical, but what practical uses are there for functions? We can use them to establish a connection between the value of one physical quantity and another. For example, through experiments or theoretical considerations we can determine a function f(p) that links the air pressure p to the air density D. It would allow us to insert any value for the air pressure p and calculate the corresponding value for the air density D, which can be quite useful. Or consider a function f(v) that establishes a connection between the velocity v of an object and the frictional forces F it experiences. This is extremely helpful when trying to determine the trajectory of the object, yet another function f(t) that specifies the link between the elapsed time t and the position x of the object. Just to give you one example of this, the function:

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connects the elapsed time t (in seconds) with the corresponding height h (in meters) for an object that is dropped from a 22 m tall tower. According to this function, the object will have reached the following height after t = 1 s:

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Insert any value for t and the function produces the object’s location at that time. In this case we are particularly interested in the root. For which value of the independent variable t does the function return the value zero? In other words: after what time does the object reach the ground? We could try inserting several values for t and hope that we find the right one. A more promising approach is setting up the equation f(x) = 0 and solving for x. This requires some knowledge in algebra.

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Subtracting 22 on both sides leads to:

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Divide by -4.91:

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And apply the square root:

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For this value of t the function f(t) becomes zero (due to inevitable rounding errors, not perfectly though). The rounding errors are also why I switched from the “is equal to”-sign = to the “is approximately equal to”-sign ≈. You should do the same in calculations whenever rounding a value.

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Graph of f(t) = -4.91t² + 22

As you can see, functions are indeed quite useful. If you had trouble understanding the algebra that led to the root t = 2.12, consider reading my free e-book “Algebra – The Very Basics” before continuing with functions.

This was an excerpt from my e-book Math Shorts – Exponential and Trigonometric Functions


New Kindle Release: Math Shorts – Exponential and Trigonometric Functions

I’m on a roll here … another math book, comin’ right up … I’m happy to announce that today I’m expanding my “Math Shorts” series with the latest release “Math Shorts – Exponential and Trigonometric Functions”. This time it’s pre-calculus and thus serves as a bridge between my permanently free e-books “Algebra – The Very Basics” and “Math Shorts – Derivatives”. Without further ado, here’s the blurb and table of contents (click cover to get to the product page on Amazon):



Before delving into the exciting fields of calculus and mathematical physics, it is necessary to gain an in-depth understanding of functions. In this book you will get to know two of the most fundamental function classes intimately: the exponential and trigonometric functions. You will learn how to visualize the graph from the equation, how to set up the function from conditions for real-world applications, how to find the roots, and much more. While prior knowledge in linear and quadratic functions is helpful, it is not necessary for understanding the contents of the book as all the core concepts are developed during the discussion and demonstrated using plenty of examples. The book also contains problems along with detailed solutions to each section. So except for the very basics of algebra, no prior knowledge is required.

Once done, you can continue your journey into mathematics, from the basics all the way to differential equations, by following the “Math Shorts” series, with the recommended reading being “Math Shorts – Derivatives” upon completion of this book. From the author of “Great Formulas Explained” and “Statistical Snacks”, here’s another down-to-earth guide to the joys of mathematics.

Table of contents:

1. Exponential Functions
1.1. Definition
1.2. Exercises
1.3. Basics Continued
1.4. Exercises
1.5. A More General Form
1.6. Exercises

2. Trigonometric Functions
2.1. The Sine Function
2.2. Exercises
2.3. The Cosine Function
2.4. Exercises
2.5. Roots
2.6. Exercises
2.7. Sine Squared And More
2.8. The Tangent Function
2.9. Exercises

3. Solutions to the Problems

Have fun reading!