17.2e Operations on Functions – Composition of Functions
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17.2e Operations on Functions – Composition of Functions

September 25, 2019


In this video we’re going to take a look at
an important fifth operation that we do on functions, and this fifth operation gets used quite frequently and it’s called a composition of functions;which simply means we put functions into functions. Visually what this means is we’ll have some function machine here which we’ll call f(x), and than another function machine lined up next to it which we’ll just call g(x) for now. And the idea here is now we’ll now plug in the a number two and f(x), we’ll get three out, the arrow doesn’t point the right spot. But than that three is automatically put into the next function, and maybe we get five out. That’s the idea of a function in function is there’s the super
function machine that does both. The super function machine when you put both two in will jump right to the final answer of five and the notation we use for this is f(g)(x). Another way
to think about of f(g)(x) is sometimes so much useful, in fact I do this on every problem to start off with. If we see f(g)(x) we’ll write f(g)(x) were I got the g inside the f function. That
means the same thing and in fact its more useful for us. That means the same thing in fact
it is probably more useful for us. So if you ever see the open circle notation, notice
it’s an open circle, it does not mean multiply it means compos.That open circle notation immediately write it with the second function inside the first, f(g)(x). And so that’s
what we’re trying to do is we’re plugging a number in. Maybe a negative three into a function, getting seven out, and then it’s immediately plugged in to the next function. Maybe giving us a
two out of these function immediately plugged to the next function. With numbers what this means is we will evaluate the inside function order of operation that is first function
and put the resolute in the outside function. This was what I was taking about with the
function machine. It puts the result immediately into another function. With variables it’s
much more the same idea, we put the inside function in for the variable in the outside
function. So let’s take a look at some examples were we do just that. We put one function
into another function and do an composition. Here in example one we’re given that f of
x is equal to the square root of six. G(x) is actually x plus three. We’re asked to find f(g)(7). Again, when we see that open dot we know it’s that’s composition. Let’s rewrite it as f(g)(x) or as seven sorry. G(7) and we need to start with the inside function.
Get a resolute and it would be plugged into the outside function. So starting with the
inside function, order of operations inside parentheses we need to know what g(7) is. Well, that means we go to replace the variable with seven, seven plus three is ten. Now that results that ten is going inside the f function. We’re going to find f(10), f of the result,
which means in f we see the x we’re going to replace it, square root of x plus six is
now the square root of ten plus six. One resolute immediately goes in the next function and
then we simplify. Ten plus six, we have the square root of sixteen, which is just four. So seven went into g, it became a ten, which went in to f it became a four. What’s interesting to know is that if
we switched the order we do get a different result. The orders very important. If we were to be asked to find g(7) this means we’re going to first figure out what f(7) is and
plug that result into g. So let’s find out what f(7) is, doing f(7) first. Let’s
get rid of the marking up here so I can match my colors. F(7) that means we’re going to
replace the x with seven in f, so we have seven plus six, and the seven plus six gives us the square root of thirteen. So f(7) is the square root of thirteen that’s what going be plugged into the g function. Inside the g we’re going to see the square root of thirteen replacing the variable. So g(x) plus three. It’s now going to be the square root of thirteen plus three and we really can’t simplify the square root of thirteen, so we just have the square root of thirteen plus three. Let’s try this again with variables and see what comes out of it. We have p(x) and r(x) and we’re asked to find p[r(x)]
or p(r)(x). Remember we rewrite in it’s useful form. Well, inside the parentheses we see r(x) and we know what r(x) is so let’s replace the r(x) with what it is. We have
p of not r(x) but x plus three, and then this x plus three is inside p. It would replace the variables. Putting them in parentheses with the x plus three. So we’ll have the x plus three squared plus two times x plus three, and we have an expression which needs to be simplified. Squaring remembering there’s three parts, x squared, and then 3x and another 3x is going to give us 6x plus
three squared is nine. Distributing the two plus 2x plus six. Combining like terms we have x squared, six and two is 8x. Nine and six is fifteen, x squared plus 8x plus fifteen. Plugging one function into the other, similar how we put in an expression into a function. And if we switch the order
again it makes a difference. When we see an rp of n. Let me get rid of everything of my
function up here. When we see r(p) of n we know p of n is this x squared plus 2x.
So what we are really asking to find is r of p of n which is x squared plus 2x. Then
this x squared plus 2x needs to go into r. It needs to be replace the variable in r. Which becomes r x is plus three. Stuff plus three, x squared plus 2x, plus three. Well, that’s kinda boring there’s not really any simplifying, so I’ll just drop the parentheses. X squared plus
2x plus three is our composition of a function inside of a function. When we see the composition dot remember it is not multiply. It is a open circle. We take one result and immediately plug it in to the other function. Which means we can be taking one function and plugging it
into the other with the variables similar how we plugged in an expiration into a function.

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