HELLO. Mr. Tarrou again. Today's topic is going to be inverse functions. Ok. So we have
got some notes. We are going to try and run through a lot of examples in the next fifteen
minutes to help you out as much as I can. We are going to be checking if two functions
are actually inverses. To do that, you have to go through the composition process of doing
f of g of x and making sure that equals x and then you have to reverse that composition
and do g of f of x and make sure that is equal to x. If both of these checks work, then your
two functions that you are given are actually inverses. To find an inverse function algebraically,
though that is not the first example that we are going to take look at, you are going
to switch the x and y variables. Resolve for your variable of y. Once you get that new
y by itself, you are going to take it out and re-label it with an exponent of negative
one. That will indicate that you are done finding the inverse and uh...the equation
that you have left is the inverse that you were looking for. First though, we are going
to take a look at inverses graphically. So, I have this function here. This f of x. If
you have been watching my videos and looked at the transformation video recently, you
might recognize this graph. We have zero here. From this point if I move over four places,
the square root of four is two. And if I keep moving over nine places, the square root of
nine is three. So you might recognize this as a square root function. But the graph does
not start at the origin, instead it starts at two. So actually this equation f of x equals
the square root of x minus two. With that plus or minus inside the function there is
a left or right movement. That minus two indicates a movement was to the right. We are not doing
transformations today, but I just thought I toss that out there for you. Ok, so we are
looking at this function which is a square root function. But, eh that is ok. I just
want a function, just a graph. Now I want to talk about graphically what does inverses
look like. Well, hint is right here algebraically. You switch the x and y variables to do your
inverse algebraically. Well, that is going to be the same thing that we do with our graph
where we have an x and y axis. Graphically how you do this inverse, you reflect the graph
around the line y equals x. Reflect about the line y equals x. Y intercept of zero and
a slope of one. And when I take this graph and reflect it around the line y=x, that line
is effectively going to switch the x and y variables. So instead of my graph going along
the x axis, it is going flip over the line...reflect over the line and go along the y axis. So
this point which was (2,0) in my reflected graph is going to be (0,2). My point that
was one, two, three, four, five, (6,2) is now going to become (2,6). So you are also
again, just taking those x and y and flipping them. One, two, three, four, five, six. And
then finally my other one is (11,3). Instead of (11,3) it is going to be (3,11). 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, and then I am off the screen. Just let me draw through the points
that I do have. This red graph is the inverse of f, or the inverse of f of x. Sorry about
that. You take the graph, you reflect it around the line y=x, and you get an inverse. Now,
is it an inverse function? Well, that is what this is about. Only one to one functions,
I have this in white to match my white graph, only one to one functions will have an inverse
function. You know a function means each x only has one y. We normally check that with
a vertical line test. Well my original graph would pass my vertical line test, because
no matter where I draw my vertical line it will only cross once. Well for this function
to have an inverse function, it must also pass the horizontal line test. So no matter
where, well I am going to match this line here, but no matter where I put my horizontal
line I am only ever going to cross once. So the vertical line test is passed. They only
ever cross once. All of the horizontal lines passed the test. I cannot draw a horizontal
line and have it pass more than once. That is why when I take this graph and I reflect
it over the line y equals x I get, look, it is an inverse. If I pass my vertical lines
through here, which were my horizontal lines, I am going to pass that line test. So only
one to one functions will give you an inverse function. That is a graphical check for that.
To give you another example of this one to one idea, if I take say this is my graph.
Let's say that it looks like this. Well that is a function, right? But if I were to pass
some horizontal lines through it, here it would cross one, two, three times. This graph
kind of goes up towards y axis. If I were to reflect this around the line y equals x,
the attempt to make an inverse function would bring this area down over here. It would look
something like. This one is going to be tough. Something like this. I am not sure if this
is the best graph, but, actually I know it is not. But it is something like that. Look,
see in certain areas if I draw vertical lines and this inverse after the reflection over
the line y = x would not pass the function test. So vertical and horizontal line test.
It has to pass both of those tests for a function to have an inverse function. Ok, that is all
with pictures. Let's take a look as some algebra problems. f of x is equal to x cubed plus
2. I want to find the inverse of this equation and I am also going to drop the function notation
for now. I want to find the inverse. Take the x and y variables and switch them. So
x equals y cubed plus two. Now I want to resolve this for y. I am going to subtract two from
both sides and get x minus two equals y cubed. I need to finish solving for y. So I am going
to take the cube root of both sides and get the cube root of x minus two equals y. I have
resolved for y. So I have switched my x and y's. I have solved for y and now I going to
make sure that I denote this is an inverse function. So I have my original function and
my inverse function. Did I make any mistakes? Are these indeed, these graphs, are these
indeed inverse functions. Well let's check it. Just temporarily I am going to call this
f(x) and this one g(x). Why? Because I am trying to match the way the notes are given
in the textbook about finding or check for inverse functions. You have to do your composition
both ways. So f and g, my original function and my inverse function. Let's check it. f
of g of x. Well function f is something cubed plus two. So this is equal to something cubed
plus two. Well what is getting plugged in? g(x). Well g is the cube root of x minus two.
So inside here. I did not leave myself very much room. I plug in my cube root of x minus
two. My next example will be written a little bit bigger:) Well the power of three and the
cube root of three will cancel each other out because the radical is acting as a grouping
symbol. So all of this is being given the power of three and all of it was also originally
wrapped with a third root. They are going cancel out. We get x minus two plus two and
that comes out to be x. So my first check has worked. Now let's make sure that our second
one does. Does g of f of x come out to also be x. Well let's see. G of f of x. G is the
cube root of the cube root of x minus two. So that is going to be the cube root of something
minus two. What is that something that we are plugging in? We are plugging in f. See,
g of f and f is x cubes plus two. So x cubed plus two. Now look what is going to happen.
Two and negative two cancel out. We get the cube root of x cubed. Once again the power
of three and the cube root will cancel out...power of three and the cube root. So once I found
my inverse function, I checked it through the composition process and both times got
an answer of x. So, I did not make any mistakes and these truly are inverses. Let's work through
another example. Let's take a look at f of x, excuse me my notes are written with function
notation. y is equal to three divided by x minus two. We want to find the inverse. So
we are going to take those x and y's, swap them around and x equals three divided by
y minus two. I have to solve that for y. I am going to add two to both sides and get
x plus two is equal to three over y. I cannot solve for a variable when it is in the denominator,
so multiply it to both sides. I am going to multiply the left and right side by y to cancel
that out. We end up with y times x plus two equals three. Now I have y times something
else. I am trying to get y by itself. How do you undo multiplication? You divide. So
y is equal to, after you divide both sides by x plus two, comes out to be three divided
by x plus two. Now that I have my y alone again I have finished finding the inverse
and I am going to denote it with my exponent of negative one. As far as a check. Let's
again call this f(x) temporarily and let's call this g(x). Let's just make sure that
we didn't make any mistakes. So is f(g(x)) going to come out to be equal to x? Well f
is three divided by something minus two. That x is your something. What is going into that
empty spot? G of x. g(x) is three over x plus two. Now, have you been watching my video?
If you have, you have seen me deal with this fraction over a fraction. Or it is 3/1 over
3/(x+2). You will know that I do this little crazy pig tail notation to say flip the bottom
up and multiply. That is going to undo the division. When you divide fraction you multiply
by the reciprocal of the second fraction and that is always going to be the bottom. So
this comes out to be three over one times x plus two over three minus two. The multiplication
of three and division of three cancels out. We get x plus two minus two and indeed that
does equal x. I am going to try really quickly to finish the other form or the other composition
before my fifteen minutes are up. So does g of f of x come out to be x. Well g is three
over something plus 2. What is that something we are plugging in? Three over x minus two.
What do we have here? Negative two and positive two will cancel out. We get three over three
over x. That bottom fraction, when you divide fractions...you can't divide fractions you
multiply by the reciprocal, and it becomes three over one times x over three. The three's
cancel out. Badda Bing, Badda Boom BAM!!! There you go. A couple of examples and some
definitions, and how to do inverse functions. Thank you very very much. WHOOO!