What is function notation? How do we use it? Click here to watch our video on this advanced algebraic concept, and learn all you need to know!
What Is Function Notation?
Another topic in advanced algebra is function notation. So first of all, I’ll say, the value of any algebraic expression of x depends on the value of x itself.
So for example, here are four different expressions. We could plug in different values of x and of course any input value of x would produce different outputs in each of these four expressions.
And of course with x–with the expression 1 over x–we can’t plug in 0 but we can plug in any other number on the other line. For the other three we could plug in any number on the number line. We can plug in positive numbers, negative numbers, or fractions. And of course we’d get all kinds of different outputs if we plugged in different inputs. So, changing the value of x, of course, changes the value of the expression.
Function notation is a way to formalize this process.
A function is like a machine that takes an input and assigns it to an output. Different inputs give the same or different outputs. For the purpose of the test, a function is always associated with an algebraic expression. The input is the value of x we are going to plug in and the output is the value of the expression once we plug in that value of x.
So function notation, for a function with the name f, we write it this way. We write f, and then we write the x in parentheses. This is read f of x. And this is not multiplication. We are not taking a number f and multiplying by the number x. F is technically what is called an operator it is performing an operation on x.
By this notation, we are saying that x is the input to the function f. Suppose f is a function that has x as its input. Plug this value of x into the expression x squared + 4, and give the value of that expression as its output. In function notation, we would write that as f of x = x squared + 4. In other words this is the rule that says give me any value of x and I will take that value of x, I will square and add 4.
That rule is formalized in this function f of x = x squared + 4. Suppose we take this function and plug a value, say x = 3, into it. We would write f of 3 = to denote that we are plugging 3 as the input into the function. To find the output we would simply plug x = 3 into the expression. So f of 3 would equal we’ll plug in the 3’s.
It will be 3 squared + 4, 9 + 4 which is 13. So the input of 3 has an output of 13.
Practice Problem One
Here’s some practice problems. Given this function, evaluate the following. You can pause the video and then we’ll talk about this.
Okay, so when we plug in f of x = 0, then of the first two terms is 0, we just got the -21. So that could be -21, f of 0 has an output of -21. F of 3, when we plug that in we get 9 + 12- 21, all that cancels, and we wind up getting 0. So f of 3 = 0, that is one of the roots of the function, one of the values that makes the function = 0. When we plug in f = -1, we get 1- 4- 21, add all this up, we get -24.
And so this is the value of the function.
Practice Problem Two
Another practice problem. Given the function f of x = x squared + 4x — 21 (the same function we had last time), find the values of x that will satisfy the equation f of x=24. In other words what inputs would give us an output of 24.
Pause the video and work on this on your own.
Okay, lets think about this. We are gonna say that that expression = 24. So really we could ignore the f of x. What we really have is just a quadratic equation. So of course the way we solve a quadratic equation, we get everything on one side equal to 0. Then we factor it. This one can be factored into x-5 x + 9. We get our two equations and we solve, so x = 5 and x = -9, these are the two solutions we get.
And so what this means is that f of 5 and f of -9 both = 24. Those are the two inputs that have an output of 24. The test will always give us the expression the function equals. The test will not expect us to find an unknown expression. But the test could give us an expression including an unknown value, and ask us about it.
Practice Problem Three
So this is a very test-like practice problem here. Given the function, f of x = x squared + kx + 4, there’s a variable that appears. An unknown value k that appears in the function. And then the question is gonna say find the value of k, if f of 2 = 18. This is very typical of a function notation question on the test.
Pause the video and then we’ll talk about this.
So, when we plug 2 into that expression, we know that that expression equals 18, so we plug it in and we simplify a bit. We get the 2 squared + 4, that’s 8. Subtract the 8 from both sides, so 2k = 10, divide by 2, we get k = 5.
All we did really was plug in the value of 2 and set it equal to 18 and then solve for k. So far, we have seen many examples of functions with quadratic expressions. Not all functions are quadratic functions, but these are quite common on the test. Linear functions are quite easy and the test can ask far fewer questions about them.
How Far Do We Go?
Many other categories of functions are harder, and quickly get into math well beyond the test. For example, trigonometric functions or something like that. These would never appear on the test. They are too advanced. The test may ask about slightly more difficult functions.
So you may see questions by functions that look like this, on the very hardest math equations on the test. So far we have talked about plugging numbers into a function, but we can also plug out algebraic expressions into functions. In the algebraic rule for a given function, x represents any input. So, what we’re plugging in for x doesn’t have to be a number, it can be a expression.
X as an All-Purpose Place Holder
Rather than replacing x with a number, we could plug in an expression and then we would replace every x in the equation with that expression. This can be conceptually difficult. Take a simple function such as f of x = 3x- 7. So this is a rule that says, give me any input, I’ll multiply it by 3, and then I’ll subtract 7, and then will be the output.
Really, writing it that way, we can write it in this form f of empty box = 3 times empty box- 7. The understanding is that we could put anything we want in to that empty box, as long as we put the same thing into the empty box on the other side. We could put a number into that empty box, we could put an expression in that empty box.
And so really what’s going on, the x of the function equation is an all-purpose holder for anything–a number, an algebraic expression, or another function. So we have to expand our understanding of what it is, what x means. What it can hold as a variable. It can hold a number but it can also hold an algebraic expression or even another function.
For example, let’s stick with this function. A nice simple linear function f of x = 3x-7, then f of 4x = well what would that mean? That means we are gonna multiply 4x by 3 and subtract 7. So we simplify this. We get 12x- 7. F of 2x- 5 we’re gonna plug 2x- 5, we’re gonna replace the x in the function equation with 2x — 5.
And so we’ll get this. Then just multiply out, simplify, and we get 6x -22. We can even plug a quadratic expression into this function. So we’re just going to multiply that quadratic expression by 3 and subtract 7. Multiplying everything out and we get 3x squared- 9x + 15 + 5.
Practice Problem Four
So here is the practice problem. Given this function x squared — 2x — 1. Find f of x squared + 3. So pause the video and then we will discuss this.
Okay, this is tricky. Every x in that original function expression, x squared- 2x- 1, every x is going to be replaced by x squared + 3. So, what we’re gonna get is x squared + 3 squared -2 times x squared + 3- 1.
Now that first expression, that’s the square of a binomial, the square of a sum, so we can use our square of a sum formula. Remember the square of the sum formula from earlier in our discussion of algebraic expressions. And so that’s just gonna be x squared squared + 2 times x squared times 3 + 3 squared.
That’s gonna be how we square x squared + 3, then we simplify all that. We get x to the 4th + 6x squared + 9 that result from squaring the first terms, then we distribute the -2 we get -2x squared- 6 and then -1. Collect all the like terms and we get x4 + 4x squared + 2 and that is the result in expression.
From plugging in the expression x squared + 3 in to this function.
In summary, the notation f of x = expression means x is the input. We plugged the value of x into the expression and the resultant value of the expression is the output. So it’s a formal way to pair inputs with outputs, that’s what a function is. We can also plug expressions into the function as the input.
In the formula of the function, we replace every x with the expression that we are plugging in. And you definitely will see problems like that on the more advanced math questions on the test.