Bam! Mr. Tarrou In this Calculus lesson we
are going to be doing a lot of review which
will be nice. We are actually.... A lot of
what we are learning in Calculus is review.
Like we first started talking about derivatives
and really just went all the way back to algebra
1 and started with finding slope. So we are
going to be reviewing inverse functions with
you. We are going to be using the Definition
of Inverse Function to validate that two functions
are inverses of themselves. Now I keep saying
function so we will be talking about domain
and range possibly needing to restrict the
domain of these functions so that they are
inverses functions and not inverse relations.
We are going to be... I am going to give you
a somewhat advanced questions and we are going
to work through finding the inverse of this
given function. Then we are ultimately going
to get into with that calculus finding the
derivative of inverse functions walking through
2 examples. The first one kind of showing
you how the formula works. And it will be
overly detailed. And then one last example
to finish up the video. So the inverse functions.
Well we know that, or hopefully remember that,
if we were given a function and asked to find
it's inverse as we are going to do in our
first big example is we take the x and y values..
we switch places... and we resolve for y.
So we have just a linear function y=2x+4.
And just pretending like I don't know how
to graph a line with just a y intercept and
a slope, I have got a table of values. There
is a reason for that. I have picked the values
of -1, 0, 1, and 2 and found our y values.
I plotted my function right here in white.
We have a y intercept of 4 and a slope of
2. And I have picked out one particular point
of (1,6). Then I have taken my x and y variables
and swapped places and then I have resolved
this second equation for y. So subtracting
both sides by 4 and dividing by 2, we get
y with an exponent of -1... which when we
are talking about inverse functions it does
not mean write 1/y. It is an inverse function.
It works in the reverse order. You know...
like trig functions. You put an angle in and
you get the sides of a triangle. With inverse
trig function, as you put in the sides of
the triangle and you get out this angle measure.
So the inverse of y, with the y being the
original function of 2x+4, is 1/2x-2. And
again I have done a table of values. My x
has a multiplication of 1/2 so I am for convenience
purposes with my table using even values.
So 2, 4, and 6, and getting the y values of
-1, 0, and 1. Then we have this, if you can
see the color of this green line right here,
is the inverse. Now a couple of things I want
to point out here just in case you forgot
it from algebra, is if you look at my table
of values I have (-1,2) where with the inverse
function I have (2,-1). I have the original
point of (0,4) changing to a point of (4,0).
Then finally (1,6) is of course being (6,1).
So not only are we seeing that we take the
x and y variables and we switch places, but
it happens with the actually coordinates in
the t-tables. So what was the domain of my
original function is becoming the range of
my inverse function. And then graphically
how that looks is we have this line of y=x.
It makes sense because we are taking those
x and y values and switching places, that
there is a reflection on this line where y
and x are equal. And again, that is allowing
the domain of the original function become
the range of the inverse function. You get
this reflection over that line of y=x. That
is the graphical way of identifying inverse
functions. So in summary inverses are reflections
over the line y=x. We notice that the domain
and the range values switch places. And we
want to put this up. This is what we are going
to focus on in the video. We can see here
in this simple linear functions, the derivative
which of course is... at least the first derivative...
is the slope of the original function... So
the derivative of function f at (c,d) is the
reciprocal of the derivative of the inverse
function at (d,c), switching again those domain
and range values. Now these are linear functions
so this slope of these functions is going
to remain constant. So I am pointing out this
(1,6) and this (6,1) because I know where
the end of the video is going. Indeed as you
look at these two linear functions we can
see that the slope of the original function
is 2 and the slope of the inverse function
is 1/2. Those are reciprocals. And for these
two linear functions the slopes are the reciprocals
everywhere in the domain. But of course with
Calculus and the first derivative that has
allowed us to find the slope of curves at
a given point. So I am just saying that if
you are at a point in one function and you
take the domain and range and swap them to
find the point on the inverse function, that
the slopes are going to be reciprocals of
themselves. Again, that becomes a big deal
in our last two examples. The Definition of
Inverse Functions: A function g is the inverse
of function f if f of g of x, so you do that
composition of functions, is equal to x when
you are done simplifying and g(f(x)) is equal
to x as well. So you set up the composition,
simplify, and if both times f(g(x)) and g(f(x))
come out to be equal to x for each x in the
domain of g and for each x in the domain of
f then f & g are inverses. Just for clarity
purposes if you needed it, I went ahead and
rewrote that expression as f of the inverse
of f of x. That inverse function is getting
put into the original function and it simplifies
to x. Then the original function getting put
into the inverse function also, also not or,
has to simplify to be x. So our first example
is going to be just using the Definition of
Inverse Functions to test whether two given
functions are inverses. We are going to do
that right now. Show that functions f and
g are inverse functions both analytically
and graphically. Now I just erased part of
my example there which I will put back up
in a second. The reason I wanted to do that
is I want to start the problem graphically
and again emphasize this restricting, some
times we need to restrict the domains of these
functions to that they are inverse functions
and not just relations. So we are going to
focus on function g first for just a moment,
even though I have kind of listed that as
the second function. But function g has got
one squared terms and it is on the x. So it
is a parabola opening, with that positive
coefficient, up. So this vertical parabola
has a vertical stretch of 4 and it has a shift
down of 1. I know my vertex is going to be
at (0,-1). Then with the vertical stretch
either I can talk about transformations which
I just hinted to, or let's just plug in a
value of 1. One squared is 1, 4 times 1 is
4, and then 4 minus 1 is equal to 3. So we
have a point of (1,3). Then this being a basic
polynomial with the x having an even power
is going to pass the test for being an even
function. That means that is symmetric to
the y-axis meaning if I plug in opposite values
x and -x I get the same y values. Or I could
just plug in -1, -1 squared is 1, 1 times
4 is 4, and 4-1 is again 3. So yes indeed
there is some vertical symmetry, or at least
symmetry to the y axis. The parabola is going
to look something like this. And a parabola
is a function. But I want to verify that these
are inverse functions, that they are inverse
functions to themselves. So if I take just
my graph right now and I reflect it over the
line y=x, instead of going along the y axis
it is run along the x axis. It would fail
the vertical line test. So while my g(x) is
a function, f(x) if it were its inverse it
would be an inverse relation. We want to restrict
these to be inverse functions. So that part
that I erased on my problem said that for
my parabola that the x has to be greater than
or equal to zero. That is not because of any
issues with graphing the parabola, but to
force these to be inverse functions not just
inverse relations. So there we go, there is
g(x). Understanding that the domain of one
function becomes the range of the other, I
could just write these two points down or
even maybe a third and just swap the x and
y coordinates and sketch my inverse function
or f(x) immediately. But to talk about transformations
for a second. f(x) is equal to 1/2, I can
pull that 1/2 out front, times the square
root of x+1. So now this function has a vertical
compression of a magnitude of 1/2. This plus
one, now this +1 is inside the math function
not at the tail end... a constant basically
all by itself, so that is going to be a shift
to the left with a compression of 1/2. So
my graph of f(x) is going to look something
like this. And we can see... I am going to
need to talk over the air conditioner for
a second before I can turn that off.... that
f(x) and g(x) as good as I can draw this on
a chalk board are showing us a reflection
over this line again of y=x. So that is our
graphical verification that these are inverse
functions. Now graphs are just that, they
are approximations. Maybe with the tables
and our calculators we can verify it more
exactly. But let's do this analytically which
is a bit more reliable. So I am going to do
the first half of this check and you are going
to do the second half. Or you can pause the
video and do both on your own. We have f(g(x)).
We are going to validate that this is equal
to just simply x. So we are going to take
function g and put into function f. So we
have the square root of something plus 1 over
2. That something that we are plugging in
for x is g. So it is going to be 4x^2-1. Ok,
so now we have the composition set up. There
is nothing in front of these parenthesis.
There are no exponents, or denominator...
There is a denominator of 2, but I am talking
about with the grouping symbol of the square
root symbol. So this negative one and positive
one are going to just simply cancel out. We
have the square root, and a broken piece of
chalk... the square root of 4x^2. The 4 is
a perfect square, the x^2 is a perfect square,
and this is just simply one term. So we can
square root the term. The square root of 4
is 2. The square root of x^2 is equal to x,
and then over 2. That simplifies to x. That
is the first half of the check. We need to
go through as well and validate 
does g(f(x)) equal x. Now you are going to
set that up on your own. One thing by the
way. When you take an even power of a value
that is also raised to that same even power,
you are supposed to need to wrap that in absolute
value symbols. So this could possible look
like 2 divided by 2 is 1 and just have a final
answer of the absolute value of x. But we
have a restriction where x must be positive.
So there is no need to write those absolute
value symbols. We are given that it is positive
as we do the problem. I am going to reveal
the answer, but run through it first and make
sure that you can do it yourself. nanananana...
Alright. Checks out both ways. g(f(x)) and
f(g(x)), they both come out to be x so these
are inverse functions. Let's go on to talking
about the existence of inverse functions.
Put into words what I have said verbally,
show you the steps of finding an inverse functions
and do one example of here is a function,
now find its inverse. We will do it a step
further than maybe we did in Algebra 2 or
Precalculus. It will involve a little bit
of practice with derivatives to check for
that 1 to 1 nature that we need to have an
inverse function. The Existence of an Inverse
Function. Well a function has to have....
For a function to have an inverse function
it must be 1 to 1. Now it has to first be
a function, that is why I keep saying function:)
But a 1 to 1 function, hopefully you remember,
and if not let's review it. That means this
function here that is increasing as we go
to the right would pass the vertical line
test. No matter where I put a vertical line
it would only cross once indicating that it
is a function. Then the 1 to 1 part means
that it would also pass the horizontal line
test. Where... Everywhere I could draw a horizontal
line it would only cross once. So that is
1 to 1 at least as a graphical check which
is what you would have done in Algebra or
Geometry. Now we are into Calculus right?
So we are going to be doing this. We need
to know how to use our graphing calculators
as well. So a visual inspection for a 1 to
1 check could certainly be sufficient. But
we can check this with derivatives as well.
If f is strictly monotonic, that means basically
increasing as you go to the right or always
decreasing. It does not change direction.
If f is strictly monotonic on the entire domain...
Let's stop there for a second... The previous
problem that I just did about checking whether
these to functions are inverses both graphically
and analytically. I first graphed that parabola
right. It was decreasing from negative infinity
to zero where we found the vertex of the parabola,
then it started to increase. That was not
a monotonic function. It was decreasing and
then there was a place where the graph changed
directions and started to increase or have
a positive slope. So we had to restrict that
domain from the vertex to just part of the
function in that particular case where it
was increasing. Not that it always needs to
be just increasing, but it does need to be
monotonic. So we restricted the domain of
that parabola such that it was monotonic.
...then it is 1 to 1 and therefore it has
an inverse function. Now as far as finding
out, or determining, if a function is monotonic,
we can type that function into our calculator
and do a visual inspection. But to review
and strengthen our Calculus skills for the
example I do with you, and you are going to
help me with by practicing the problem yourself,
we are going to check that monotonic situation
by looking at the first derivative. Basically
like doing a first derivative test. Steps
for finding an inverse function. Well, we
need to check if it is 1 to 1 visually or
with your first derivative. We are going to
swap the x and y variables again as you saw
me do with this linear function. We are going
to resolve after swapping the x and y variables,
resolve for y and then label that as the inverse
function... whether it is y^-1 or f^-1(x).
We read that as the inverse of f(x) or the
inverse of y. Determine the domain of the
inverse function. You know determining the
domain of a function is much easier quite
often than determining the range. So it says
here we are going to determine the domain
of the inverse function which is the range
of the original function. We can take that
original function f(x) identify its domain,
then find the inverse and find the domain
of the inverse function which is the range
of the original function. Now determining
domain, your trig functions are a little bit
more complicated with maybe where the vertical
asymptotes. Say for tangent function, they
may be a little bit hidden and not obvious
from the equation unless you are really know
your unit circle and your trig functions,
of course we should. But if you are looking
at a pure algebraic equation, we can be very
cautious of restrictions in the domain if
the dependent variable... the domain... or
the x variable is inside an even root or in
the denominator of a rational function. It
does not guarantee that there is a restriction...
like if the variable is in the denominator...
but it probably will have one. Then of course
if you want to take it that one extra step
and make sure that you answers are inverses
of each other you can check your definition
of inverse functions and make sure that both
of these expressions comes out to be X. We
are not going to do this last step in our
example of finding an inverse function. But
we are going to do the other four steps. Maybe
again a bit more in-depth then maybe you would.
It would be easier to have the aid of a calculator,
but we are going to practice our skills of
derivatives as we do this. So let's get to
that example of finding an inverse function.
Find the inverse function of f and state the
domain and range of the original function
f and the inverse of f. Our original function
is f(x) is equal to 2x divided by the square
root of x^2+5. Now with use of our graphing
calculators which is something definitely
need to practice we can type this into our
calculator, analyze the graph, and by analyzing
the graph determine the domain and there is
some horizontal asymptotes going on here.
So I would have to recognize that asymptotic
behavior, but we should also be able to identify
the range. We should also be able to just
look at the graph and determine if it is monotonic
increasing or decreasing. In other words is
it 1 to 1 and thus has an inverse function.
But you know that is fairly straight forward.
Hopefully if you know how to use your graphing
calculator. But I want allow you the opportunity
to remind yourself, restrengthen your derivative
skills in finding horizontal asymptotes of
irrational functions. So I am going to step
out and reveal these solutions one step at
a time. Maybe pause the video, find that first
derivative, determine if it is monotonic increasing
or decreasing based on the first derivative,
and then if you remember how to find those
horizontal asymptotes find the range of this
function as well. We can again find the domain
of the inverse function which is a little
bit easier which will be the range of this
original function. But I just want to review
some skills with you as we go through the
video. Just quickly run through this. We have
a domain... now our denominator... we have
variables that are in the denominator. We
cannot divide by zero, nor can we take an
even root of a negative number. So with a
combination of those two we set x^2+5 is greater
than 0. We subtract both sides by 5. We have
x^2 is greater than -5. Well certainly that
we raise to an even power is going to be positive
so this will automatically be greater than
negative five. So the solution is all real
numbers. For testing whether the function
is monotonic thus being 1 to 1 we can find
the first derivative here. With the quotient
rule we have the denominator times the derivative
of the numerator minus the numerator times
the derivative of the denominator, which required
the chain rule over the denominator squared...
the 2 cancels out... we have a factor of x^2+5
to the 1/2 power and the -1/2 power. That
same factor is both terms so we take out that
lowest power of -1/2. Both of these terms
to the left and the right side of the minus
sign have a factor of 2, so that 2 can be
factored out as well. A little simplification
and we have f'(x) is equal to 10 over x^2+5
raised to the 3/2 power. So what is going
on here is again our x has an even power,
so whatever we put in for x becomes positive
with the even power, then we add by 5 keeping
it positive, and we have a positive numerator
so this expression... take a positive and
raise it to an odd power comes out to be positive...
the even root with no negative out front becomes
out to be positive.. So the first derivative
is going to be greater than zero or positive
everywhere within the domain which is all
real numbers. Thus this is monotonic increasing.
So thus it us one to one and we will have
an inverse function. Now coming over here
I am going to point something out at the end
of this that maybe would saved us a little
bit of time if we had thought of it at first.
But horizontal asymptotes, this is an irrational
function. There is a fraction bar so it is
a fraction, but then it has a square root
in it so this is an irrational function. So
we need to go through this limit process.
To find the limit as x approaches positive
infinity we multiply the numerator by 1/x
and to get that into the square root we take
that x and raise it to a second degree and
square root it. So we have 1 over the square
root of x^2 which simplifies to be x when
x>0. So this is equivalent to multiplying
by 1. When we do that we get ultimately a
limit of 2 so as x approaches infinity the
graph is going to approach the value of 2.
And to approach negative infinity we need
to have some kind of sign change to indicate
that in our limit process. We can either do
-1/x, or 1 over negative square root of x
squared as I indicated. Which is nice because
then you just put in a negative in front of
your radical and the limit as x approaches
negative infinity is going to be negative
two. So we have... I didn't mark off, maybe
I should have, the horizontal asymptotes.
But if we were to have graphed the original
function and tried to find domain and range,
and whether the function was 1 to 1 with a
graph or a graphing calculator we would have
seen this kind of graph. Just to again review
a lot of concepts that we have learned in
the past, if I were to take this function
and rotate it 180 degrees it would appear
to have symmetry about the origin. Well that
means that it is an odd function. An odd function
are where you plug in opposite values of x
and negative x and you get out opposite answers.
Well if I plug in opposite numbers into the
denominator, the negative or the opposite
will be cancelled out by the even power thus
the denominator is going to be the same...
it is always going to be the same whether
you plug in a positive or negative x. But
the numerator is just simply 2x, so yes when
you plug in opposite values into this function...
when you plug in -x because the sign of the
numerator changes and the denominator does
not, you are going to get opposite answers.
Thus this would pass the test for the being
an odd function. Now we can see that whether
you graph this with a table, or more than
likely with your graphing calculator. So we
have not found the inverse function yet, but
we have done a lot of work of finding domain,
finding whether it is 1 to 1, the extra work
of finding the range. This is me just again
reviewing concepts. We could have found the
inverse first and found the domain of the
inverse which would still give us the same
values between -2 and 2. Let's go ahead and
find that inverse function now. Got everything
cleaned up. Summarizing what we just found,
the domain, our range, and our sketch. We
are going to take these x and y variables
and swap places. So instead of y we are going
to have x is equal to 2y over the square root
of y squared plus five. Again, we want to
now solve this for y. We are going to square
both sides to get rid of this even root and
manipulate the equation to the point where
we get y in terms of x. That will be the inverse
function. Again I am going to step out and
reveal the solution one step at a time. Got
those x and y's swapped of course. That is
how we started the problem. I squared both
sides of the equation to get rid of the square
root. We multiplied both sides by y^2+5, distributed
the x^2 into the parenthesis. I moved my four
y squared over to the left with subtraction,
and the same moving 5x^2 to the right. When
I got down to this point I have y squared
is equal to -5x^2 over x^2-4. Well if I go
to get rid of this power of 2 by taking the
square root both sides with that x squared
being positive... with that x^2 term whatever
I plug in for x is going to be positive because
of the even power... then I get multiplied
by negative five. Now I cannot square root
a negative number. So in this format the numerator
is guaranteed to be negative and thus I cannot
take the square root of it. So I multiply
the numerator and denominator by -1 to get
rid of that negative that is in the numerator
and also of course that is going to take the
binomial in the denominator and swap the order.
x^2 times negative one is negative x squared,
and -4 times -1 is positive 4. Now this expression
has a pretty tight restricted domain for where..
you know... it can be positive but it is not
guaranteed to be negative like we had in the
numerator. So I go ahead and then take the
square root of both sides. I identify it as
the inverse of y. I get plus or minus the
square root of 5 x over the square root of
4-x^2. Now again you cannot take an even root
of a negative number and you cannot divide
by zero, so four minus x squared needs to
be greater than zero. We get solved for x
squared, we square root both sides, and with
inequalities you need to change the direction
of the sign for the negative answer. So we
have that negative two is less than x which
is less than 2. Right!? The domain, I just
solved for the domain of the inverse function
and it is checking out to be the same as the
range of the original function. It is all
checking out to be good. We need to deal with
this plus or minus that we have. I didn't
pick a real simple question for finding the
inverse. We have a plus or minus here. Well,
it is not both ok. We are looking for the
function which the reflection over the line
y=x which is something like this. This is
just a quick sketch. I don't know if this
really crosses or not. But we are looking
for the reflection of this original function
f(x) over the line y=x. If you use your graphing
calculator to graph both the positive and
the negative version of your answer, one of
those answers... which is the negative one...
will come down and cross through origin this
way. It will still have vertical asymptotes
of x=-2 and x=2, but it will be coming through
like this which is not your reflection over
the line y=x. We are looking for a graph that
looks like this. And the graph that is that
reflection over the line y=x is the positive
version of our answer. So our inverse function
is the square root of 5 times x over the square
root of 4-x^2. And you can check that if you
like with the Definition of Inverse Functions.
I have graphical check right here so we are
going to leave it at that. We are going to
move on to the derivative of in inverse function,
write out that theorem, and some formulas
so we can get to our last two examples. ... and
find my remote. Let's talk about a few properties
of an inverse function. If f is a function
whose domain is an interval I. If f has an
inverse function, then the following statements
are true. If f is continuous on its domain
then the inverse of f is continuous on its
domain which is the range of the original
function. If f is increasing which means that
its first derivative is positive on its domain,
then the inverse of f is increasing on its
domain... so its first derivative is going
to be positive. If f is decreasing on its
domain, then the inverse function f^-1 is
decreasing on its domain. Finally if f is
differentiable on an interval containing c
and if f prime of c is not equal to zero,
so basically the slope of that original function
is not equal to zero... because when you reflect
that over the line y=x, that horizontal tangent
line is going to become vertical which will
be an issue with trying to find the derivative,
so that derivative of f of c cannot equal
zero... then the inverse function or the inverse
of f is differentiable at f(c). Now again
that domain and range is swapping right as
you find inverses. So f is differentiable
at c, and the inverse is differentiable at
f(c). f(c) is the y value of your original
function. But when you take the inverse and
you start talking about inverse functions,
the y of the original function is the x of
the inverse function. That is why you see
that f(c) there. Ok, let's look at the rule
for the derivative of an inverse function.
This rule that allows us to do this more quickly
than actually having to find the inverse...
then having to find the first derivative to
find this slope. BAM!!! This is what the whole
video is building up to, The Derivative of
an Inverse Function. Let f be function that
is differentiable on an interval I. If f has
an inverse function which we are going to
identify as g, then g is differentiable at
any x for which the derivative of f, or f
prime, of g(x) is not equal to zero. This
is how the formula is initially given in the
book that I teach out of in class. It says
the derivative of g of x is equal to one divided
by the derivative of f, or f prime, of g of
x. Ok. I find that initially can be confusing.
So I am going to take this and just write
it in terms of the inverse function notation.
The derivative of the inverse of f of x is
equal to one over the derivative of f of the
inverse of f of x. Even that is kind of confusing.
So let's see here. Derivative is slope. So
the slope of the inverse function at its x
is equal to 1 divided by the slope... the
derivative... the slope of the original function
and we are plugging in the inverse of f of
x. It is easy for me to say. Basically the
y that we get from the inverse function. Well
why are we asking for y? Well, the x of the
inverse function is the y of the original
function. Remember we talked about that domain
and range swapping places when we find inverses.
So the y of the inverse function is the x
of the original function. Let's go back to
that original diagram that I started with
this video, those two linear functions...
which slopes do not change but just for simplicity
sakes. We have y=2x+4. The slope of the original
function was 2. We swapped the x and y, resolved
for y, and we got the inverse of y was equal
to (1/2)x-2. Well yeah, these slopes are reciprocals
of each other. These slopes are reciprocals
of each other. And this formula says that
the slope of your inverse function at its
x is equal to the reciprocal... one over the
slope of the original function... but you
are plugging in the y again of that language
of the inverse... or the y of the inverse
which is 2 was the x of the original function.
So the slope of the inverse function at its
x is equal to the reciprocal of the original
at that inverse's y value. That y value was
the x of the original function. So right here
again 1 over f' or the slope of the original
function where you are plugging in the y of
the inverse function which was the x of the
original function. I am going to really hit
on this really hard. Kind of actually do the
first example longer than I need to just to
show you that swapping of the domain and range
and how those slopes are reciprocals of each
other. Then we will do the last example very
streamlined allowing this formula to basically
really short cut the process to help us find
derivatives at particular x values of the
inverse function without finding the inverse.
Verify that f has an inverse. We are going
to assume that we are talking about inverses
like we have all video. Then use the function
f and the given real number 'a' to find the
derivate of the inverse of f of a. Ok. So
normally I just tell my students if you really
get stuck on how to do the problem, just let
the formula tell you what you need. Ok. First
thing it says is verify that f has an inverse
function. It needs to be monotonic. We are
going to find the first derivative. f prime
of x is equal to... that is a constant so
it is zero... and -3 times 3 is -9x^2. So
no matter what x is, it is going to be squared
and therefore this factor will be positive...
not times negative nine will be negative.
So the first derivative is always going to
be less than zero so this is monotonic decreasing.
a is equal to 5. So the derivative of the
inverse of f of a is equal to 1 divided by...
Well I need the derivative. Actually I just
figured that out, right. It is -9 times something
squared. Then if you did not really catch
what I was trying to point out to you in the
previous screen, then you might be going I
need the inverse of f(a). Maybe we will burn
some calories and do that. And if I want the
inverse function I am going to take the x
and y and swap them. Let's do this in orange.
So we have x... this is the extra part that
you don't really have to do. I am just trying
to tie everything together and we will streamline
the last example. This y is going to become
x. We are going to put a y here. We are going
to resolve this for y. We are going to subtract
both sides by 2, divide by negative three.
I want to turn this around too. Then we are
going to take the cube root of both sides.
Now that I have finished resolving for y,
this is my inverse function. I am not going
to check it with the composition of f(g(x))
and g(f(x)). But this is my inverse function.
Now that I have the inverse of y, I can come
up here and go... I was looking for the inverse
and that is the cube root 
of x, which I am plugging in 'a' so it is
going to be 5 minus 2 over negative 3 all
squared. This is going to come out to 1 over
negative nine times... 5-2=3 and 3 divided
by -3 is -1 squared. And the cube root of
-1 is -1. -1 squared is positive 1. Positive
one times nine gives us a final answer of
-1/9. The derivative or the slope of my inverse
function at 'a' is equal to negative one-ninth.
Now what I want to point out here and to show
what I was trying to point out in the previous
screen, I took the time to find the inverse
function and then find that value of that
inverse function at 'a' which came out to
be -1. Then we plugged that into the first
derivative. Ok. I did not actually have to
find the inverse function. There will probably
be problems in your homework where you really
will have a very difficult time finding that
inverse function. So we can stream line this
if we remember that the 'a' again was effectively...
see it going into the inverse function and
thus giving us a y value. Well that 'a' is
the domain of the inverse and it is the range
of the original function. So I can say I am
recognizing now that the 5 is actually the
y value for the original function. So if I
take this original function, let me get this
out of the way, and say 5 is equal to 2-3x^3.
Subtract both sides by 2. Divide both sides
by -3. We get -1 is equal to x^3. Then cube
root both sides and get negative one is equal
to x. The x that is there from my original
function becomes the y of my inverse function.
Look at what I have inside for this value
of the inverse of f of a. We have negative
one. So I don't need to find the inverse function
to find the slope of the inverse function
at 'a'. I just have to recognize this swap
of the domain and range is going on, and say
ok... the x for the inverse function is actually
the y of the original function. I can find
that inverse of f of 'a' a lot easier. Then
I can just plug in the negative one right
here. I think I pointed to that line, I should
have been pointing to this one. Plug in -1
here and get my answer a lot quicker. So let's
stream line that as our final answer and last
example of the video. As our last example
f(x) is x+4 over x-3 with a restricted domain
where x is greater than 3. We have an 'a'
value of 8. So we are going to find the slope
of the inverse function at the value of 8.
To stream line this again we are going to
find the derivative of f(x) to verify that
it is 1 to 1, thus the inverse function exists.
Plus we need it as a denominator down here.
We are going to set f(x) equal to 'a', solve
it for x which is the y value of that inverse
function... plug it in, get our answer and
we will be done much quicker than the last
example. So f prime of x is equal to the denominator
times the derivative of the numerator minus
the numerator times the derivative of the
denominator over the denominator squared.
So when we get done cleaning all of this up...
this is getting multiplied by 1 so that is
not going to do anything... distribute that
negative one. We have f prime of x is equal
to x minus x which is zero... negative three
minus four which is negative seven... over
x-3 squared. Now the denominator is always
going to be positive with that even power.
The numerator is negative, so our first derivative
is going to be negative always so this is
monotonic decreasing thus it is 1 to 1. Ok.
We need to find this y value here, the inverse
of f of 'a'. So we are going to let this function
equal 8. 8 is equal to x plus four over x
minus three. Multiply both sides by x-3 and
we get 8x-24 after we distribute which is
equal to x+4. Move this x over to the left
and we get 7x, add both sides by 24 and we
get 28. Divide both sides by 7, and we get
x is equal to 4. So the slope of the inverse
function at 8 is equal to 1 over... ok...
f prime, the slope of the original function.
Well that is -7 over parenthesis something
minus three squared. And again the y of our
inverse function, the f^-1(a), is the x of
our original function here which we found
to be 4. It is going to be four minus three
squared. Well 4-3 is 1. 1 squared is 1. Negative
seven over one is just negative seven. Our
final answer is -1/7. I started finalizing
the numbers and I had a little interruption.
So what we have here is the slope of the original
function at 4 was -7, so swapping the domain
and range the slope of our inverse function
at 8 is the reciprocal which is -1/7. So I
am Mr. Tarrou. BAM! Go Do Your Homework...
and I will find my remote.
