HELLO!!! Mr. Tarrou again. Same shirt, same day. Multiple recordings. I am going to talk
about composition of functions. Not just combining them, not just adding subtracting, multiplying,
or dividing but actually a composition of functions. Composition notation is f open
dot g of x, or... This is the notation my textbook uses a lot but you may also see it
as f of g of x. I like this notation better because really more clearly explains what
is happening. We are going to take function g and put in into f. With this notation you
are forced to read right to left which is not how normally read. So, let's talk about
domain. Basically this is fancy language to say that when you talk about composition functions
you have to check the domain of the individual pieces and the final answer. If x is not in
the domain of g, if x makes g undefined it must not be in the domain of your final answer
f(g(x)), even if your final simplified answer makes it look like it will be ok. Any x for
which g of x is not in the domain of f, so you take x and plug into the function of g
and it works just fine, but then you plug into the domain of f and that comes out to
be undefined then again that must be thrown out for the domain values of f(g(x)). This
is a pretty simple idea as long as you are comfortable with function notation. Let's
see what it looks like. f of x is going to be two x minus three and g of x is going to
be five x squared minus one. So, let's first find out what f of g is, or f of g of x. We
want to start off with the outside function, the overall pattern that we are following
is function f. What is function f? Function f is not 2x minus three. No, no, no. X is
a place where plug something in, so I like to say that f or the outside function is two
times something minus three. I don't care what I am plugging in. X is some empty space
where you are going to plug in something else, that something else is g. See it is f of g
of x and g of x is five x squared minus one. So, five x squared minus one. Clean it all
up, work it out, simplify, blah blah blah. Two and five is ten x squared plus, two and
negative one, is negative two minus three is negative five. What about your domain and
range? No fractions, no radicals, no fractions no radicals, actually not radical but even
roots, and no fractions or even roots so domain is all real numbers. Lets try this the other
way and see what happens. What is g of f, well that is g of f of x. We are just going
to substitute in the opposite direction. What is g? Well I am glad you asked! It is five
times something squared minus one. So it is five times something squared minus one. What
is that something you ask? It is f of x and f of x is two x minus three. Now I am going
to run out of of space so I am going to do this off the top of my head and hopefully
I am right, or I can put this off to the side and distribute this out and make sure you
get all three terms not two. This is going to be five times four x squared minus twelve
x plus nine minus one. If you you did not get that, go off to the side of your paper,
write (2x-3) twice and distribute that together and you will go OHHH I know where that came
from. Now distribute the five through and you get 20 x squared minus 60x plus 45 minus
1. Combine like terms, clearly that is going to be 44. And BAM!!! I got the composition
working both ways. Again F has a domain of all real numbers and g(f(x)) is defined for
all real numbers. What if, I am going to take this away so make sure you have a copy, we
take a look at...or I ask you, "What is f of g of 2?" I don't know why I wrote that
in parenthesis, I don't need to. I just worked out what f(g(x)) was, it was ten x squared
minus five. So if I take x out and plug in 2, what am i going to do? I am going to take
out the x and plug in two. Ten times two squared minus five. That is four so ten times four
is forty, and forty minus five is thirty-five. So there you go, just take out the x and plug
in the number. Badda Bing, badda boom. Ok, any questions...didn't think so:P Let's take
a look at.... If you had any questions, I wouldn't know on this video. Let's take a
look at the next example. A little bit of fraction work. We have f of x is equal to
x over x plus one and g of x is equal to three over x. I am sure you have some questions,
this stuff can get a little bit confusing. Just ask your teacher about the function notations
if you need some extra help. We want to find f of g of x, or f(g(x)). Well again, this
is your framework function f, the outside function, or the first notation. F is something
over something plus one. That something you are plugging in is three over x. Ok. I am
going to need a little bit of space so let me take out these parenthesis. They were not
doing anything except highlighting where I am going to do my substitution. Again like
the last video, if you just watched it, we are going to have a fraction over a fraction.
We need that denominator to be one fraction and then we are going to flip it up. So we
are going to get a common denominator and that common denominator is going to be x,
so this is going to be x over x. You know multiply the top and bottom by x. So three
plus x over x. Voila, three over x over three plus x over x again. And a fraction divided
by a fraction, you cannot really divide fractions you have to multiply by the reciprocal. Do
your little pig tail and flip that denominator up. Look the x's are going to cancel out.
So our final answer is going to be three over three plus x. Now let's talk about domain
for a second. This is going to be undefined when x equals negative three, so x cannot
be equal to negative three. But is that the only problem with our domain? I don't think
so. G of x is undefined at zero, and the rules say "If x is not in the domain of g, it must
not be in the domain of f of g. So x can also not be equal to zero. And negative one? Any
x for which g of x is not in the domain of f, must not be in the domain of f(g(x)). So
can x be negative one...well let's see. This becomes undefined but negative one does not
make g of x undefined which is what we are plugging in and the final equation is not
undefined at negative one, so that is going to be ok. It does not say anything about x
making f, the outside function, undefined. It is about x making g (your inside function)
undefined and x making g of x making f undefined. So maybe some of this stuff isn't quite so
simple. But we have a domain which is restricted at negative three and zero. Let's look at
the other direction. What is g of f of x, or g(f(x)). That is going to be, the framework is g so it is going to be
three over something. What is that something that we are plugging in? The something we
are plugging in is x over x plus one. And fraction over fraction again, so that is going
to get flipped up. If you want to write this because three does not look like a fraction,
it is three over one times x plus one over x. So our final answer is going to be three
x plus three over x. X cannot be equal to zero and x cannot be equal to negative on.
Check the domain of the inside function, then check the domain of the final answer. BAM!!!