(male narrator)

In this video, we will begin looking

at the composition of functions. What a composition

of function is is when we

have functions… in functions. We will use

the notation f open circle g of x

to mean f of g of x. What this means is we

plug x into the g function, and then, we plug the result

into the f function. In the other notation,

x into the g function, and the result

into the f function. Let’s take a look at some

examples where we do just this. In this problem,

we’re finding f of g of 7, which could also be written

f of g of 7. What this

is asking us to do is first plug 7

into the g function, so let’s do that. When we replace the x with 7

in the g function, we find g of 7 is equal

to 7 plus 3, which is 10. Then, what this

wants us to do is plug the result

into the f function. In other words, we plug

into the f function the answer we

came up with: 10. Replacing the x with 10

gives us the square root of 10; plus 6; or the square root

of 16…which is 4. Our final answer for the

composition f of g of 7 is 4. Let’s take a look at doing

the same type of problem, but this time switch

the order of the g and the f. This means

we do the g with the f function

inside of it. In other words, we’re gonna plug

7 first into the f function: f of 7 is equal

to the square root of x, or 7 plus 6, which is

the square root of 13 that can’t be

simplified. Now, we’re asked

to take that result and plug it

into the g function. In other words, we take g of the result, square root of 13; g is x plus 3, or the square root of 13 plus 3. Can’t simplify this expression, and so that becomes our final result. It is important to note that when we switch

the order of the composition, we do get

a very different result. We can also do composition

at just an x value. Here, we plug…

take p of r of x. This means we take the r of x,

since we can’t evaluate it, and plug it

into the p function. In other words, we take all

of the r of x and plug it in for the p function

for each of the x’s. In the p function,

we’ll see r of x–or x plus 3– which means we have x,

which is now x plus 3; plugging the r function in;

squared; plus 2 times x, which is the r function

x plus 3. Squaring will give us x squared;

the product twice, 3x and 3x is 6x;

and 3 squared is 9. Distributing gives us

2x plus 6, and then, we can finish

by combining like terms to get x squared;

plus 8x; plus 15; for our function:

p of r of x. In other words, we stick

the r of x into the p function. When we switch

the order, now we’re asking

to take the p of x function and plug it

into the r function. The r function–r of x–

becomes r of x squared plus 2x, where x is now

x squared plus 2x; and then, we add 3

to the end. With no simplifying to do,

our final answer is x squared, plus 2x, plus 3.

So so lost!

Yo Tyrone, I happy to lern da math, but next time, put in in da mixtape with da trap song so ayy can lmao