Algebra of Functions – Composition
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Algebra of Functions – Composition

August 24, 2019


(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.

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