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Another topic in advanced algebra is function notation. The first of all I’ll say the value of any algebraic expression of x depends on the value of x itself. For example here are four different expressions. We could plugin different values of x. And of course any input value of x would produce different outputs in each of these four expressions.
That of course with x eq, with the expression 1 over x. We can’t plugin 0 to that, but we could plugin any other number on the other line. For the other three we could plugin any number on the number line. We can plug in positive numbers, negative numbers, fractions and of course we 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 plugin, and the output is the value of the expression once we plugin that value of x.
The function notation, for a function with the name f we write it this way. We write f and then we write the x in parenthesis. This is read, f of x, and this is not multiplication. We’re not taking a number f, and multiplying by the number x. F is technically what’s called an operator. It is performing an operation on x.
By this notation, we are saying that f is the input to the function f. Suppose f is a function that has x as its input and plugs this value of x into the expression x squared plus 4 and gives the value of that expression as its output. In function notation, we would write that as f of x equals x squared plus 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 it and add 4.
That, that rule is formalized in this function. F of x equals x squared plus 4. Suppose we take this function and plug a value, say x equals 3 into it. We would write f of 3 to denote that we’re plugging 3 as the input into the function. To find the output, we’d simply plug x equals 3 into the expression. So f of 3 would equal.
We plugin the 3’s. So it would be 3 squared plus 4. 9 plus 4, which is 13. So the input of 3 has an output of 13. Here’s a practice problem. Given this function, evaluate the following.
You can pause the video and then we’ll talk about these. Okay, so when we plugin x equals, fx equals zero, then of course, the first two terms is zero. We just kept the negative 21. So that’s gonna be negative 21, f of 0 has an output of negative of 21. F of 3.
When we plug that in, we get 9 plus 12 minus 21. All that cancels and we wind up getting 0. So f of 3 equals 0. That is one of the roots of the function, one of the values that makes the function equal 0. When we plugin x equals negative 1.
We get 1 minus 4 minus 21, when we add all this up, we get negative 24. And so this is the value of the function. Another practice problem. Given the function, f of x equals x squared plus 4x minus 21. The same function we had last time. Find the values of x that would satisfy the equation f of x equals 24.
In other words what input would give us an output of 24. Pause the video and work on this on your own. Okay, let’s think about this. We’re gonna say that, that expression equals 24, so really we can ignore the f of x, what we have is just a quadratic equation there. 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 minus 9, x minus 5, x plus 9. We get our two equations and we solve. So x equals 5, and x equals negative 9. These are the two solutions we get. And so what this means.
Is that f of 5 and f of negative 9 both equal 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 good, could give us an expression, including an unknown value, and ask us about it.
So this is a very test-like practice problem here. Given the function f of x equals x squared plus kx plus 4. So 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 equals 18. This is very typical of a, of a function notation question on the test.
So post the video and then we’ll talk about this. Okay, so when we plug 2 into that expression, we know that, that expression equals 18. So we plug it in we simplify a bit, we get the 2 squared plus 4, that’s 8, subtract 8 from both sides, so 2k equals 10, divide by 2, we get k equals 5.
So all we did really was plugin 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 expression and 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.
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, these would never appear on the test. They’re too advanced. The test may ask about slightly more difficult functions, so you may see questions.
About functions that look like this on the very hardest math questions on the test. So far, we have talked about plugging numbers into a function. But we can also plug 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 an expression.
Rather than replacing x with a number, we could plugin 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 equals negative 3x minus 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 that will be the output.
Really. Writing it that way, we could write it in this, this form. F of empty box equals 3 times empty box minus 7, with the understanding that we could put anything we want into 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 into 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 equals 3x minus 7, then f of 4x, well what would that mean? That’s means we’re gonna multiply 4x by 3 and subtract 7. So we simplify this. We get 12x minus 7. F of 2x minus 5.
We’re gonna plug 2x minus 5 we’re gonna replace the x in the function equation with 2x minus 5. And so we’ll get this. Then, just multiply out, simplify and we get 6x minus 22. We could even plug a quadratic expression into this function. So, we’re just gonna multiply that quadratic expression by 3 and subtract 7.
Multiplying everything out, and we get 3x squared minus 9x plus 15, plus 5. So, here’s a practice problem. Given this function, x squared minus 2x minus 1, find f of x squared plus 3. So pause the video, and then we’ll discuss this. Okay, this is tricky.
Every x in that original function expression, x squared minus 2x minus 1, every x is going to be replaced by x squared plus 3. So what we’re gonna get is x squared plus 3 squared. Minus 2 times x squared plus 3 minus 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 about algebraic expressions. And so that’s just gonna be x squared, squared plus 2 times x squared times 3 plus 3 squared. That’s gonna be how we square x squared plus 3.
Then we simplify all that, we get x to the fourth plus six x squared plus 9, that results from squaring the first term, then we distribute the negative 2, we get negative 2x squared minus 6 and then a minus 1. Collect all the like terms, and we get x to the fourth plus 4x squared plus 2, and that is the resulting expression. From plugging in the expression x squared plus 3 into this function.
In summary, the notation f of x equals an expression means x is the input. We plug 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. The formula for the, in the formula 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.
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