BAM!!! Mr. Tarrou In this video we are going to define and do a couple of examples involving
the Rolle's Theorem. Hopefully I am saying that man's name correctly. It is basically
like a special case of the Mean Value Theorem. It says, let x be continuous on the closed
interval of [a,b] differentiable on the open interval of (a,b). If f(a)=f(b), then there
is at least one number c in this open interval from a to b where f prime of c is equal to
zero. Now, I have drawn sort of the bottom end of a parabola here, or just some function
with a minimum value. And, I have got it marked off where, let's call this function g, and
g(a) is equal to g(b). I am trying to draw this of course so these points are horizontal
from each other. Through these two points, it does not specifically say secant line in
this theorem, but if g(a) is equal to g(b) then if you were to draw a secant line through
those two points it would be horizontal and thus have a slope of zero. The first derivatives
of course again tell us the slope of a curved function, or any graph for that matter. So
it just says that somewhere within this interval from a to b if g(a) and g(b) are equal, then
there is a value c, an x coordinate of c, such that when you plug that value of c into
the derivative you get a slope that is equal to zero or you get a horizontal tangent line
at that point of g(c). So, let's practice...oops, I almost forgot. We are going to do two examples,
but I want to discuss why we have this statement within the theorem that it is differentiable.
That is because, and there might be other instances of this, but think of absolute value
functions. Absolute value functions, they don't have to form two linear patterns and
form a v, but they do have a sharp bend in them. So, if I say here is function f, and
I am going to make mine a straight v, we have a couple of points at a and b where f(a) is
equal to f(b). So we have that horizontal secant line. Our function, whether it goes
up or down does change value. I have mine going down and I am going to have a minimum
like I do in this function, but there is a sharp bend. Well, maybe this slope is negative
one-half and this slope is positive one-half. Or maybe this slope is negative one and positive
one. But somewhere at, let's mark this off as c again, at our minimum value we have a
sharp bend. Well you cannot find the derivative at a sharp bend, so thus with this example
even though f(a) is equal to f(b) there is going to be no derivative value, there is
going to be no place where you can plug in the value of c into the derivative and get
a slope of zero. That is because of this sharp bend and we cannot find the derivative, or
this function is not differentiable at the value of c. So we do need that smooth continuous
curve, not that continuous curve with that sharp bend because we cannot find the first
derivative or any derivative for that matter at those points. Let's look at those examples.
One is going to be algebraic and one is going to involve trig functions. WHOOO!!! So here
is our example. We have, find the x intercepts. Which means that we are going to find two
x values where the y values are equal to zero. So there is your horizontal secant line. Show
that there is a value of c withing that interval that we find in the first step where we find
the x intercepts, that contains, or where the f prime of c...where the first derivative
is equal to zero. We have f(x) is equal to x times the square root of x plus nine. So,
let's find some x intercepts. Now for x intercepts, of course you need your y value to equal zero.
That is what happens when you cross the x axis. So, zero equals x times the square root
of x plus nine. Now this is basically in factored form. It does not really look like how we
would normally expect to see it in algebra, or algebra 2, or PreCalculus for that matter.
But we do have one factor of x being multiplied by a second factor which is the square root
of x plus nine. Don't forget that square root functions, or any math function like a fraction
bar or an absolute value symbol act as a grouping symbol. So it is one thing here with the first
x, this first factor, times this second something that just happens to be a little bit more
complicated than we normally see. Now, since it is already in factored form, and there
is more than one way to get this solution, I want to just write this well either the
first factor needs to be equal to zero or the second factor which is the square root
of x plus nine has to be equal to zero. If either one of those factors equals zero, then
the entire equation is equal to zero. Well this answer is already done. My first x intercept
is zero. Over here we have the square root of x plus nine is equal to zero. You know
I am hesitating because I am sure some of you can get this answer in your head. But,
just in case the question is a bit more complicated, I want to remind you that of course we do
get rid of those square roots by squaring both sides of the equation. We get x plus
nine is equal to zero. Therefore, when you subtract both sides by nine you get x equals
negative nine. So, f(0) equals zero and f(-9) equals zero. So, somewhere within that interval
the Rolle's Theorem says that the first derivative , there will be some value of c, some x value
between negative nine and zero, there is some c value between negative nine and zero, there
is some value that when you plug it into the first derivative you will get an answer of
zero. So, let's find that first derivative and see if it is true, or if I have worked
out our problem correctly because I don't need to prove Rolle's Theorem. f(x) is equal
to x times the square root of x plus 9 to the one-half power. I don't want to do derivatives
with radical symbols. This is the product property. BAM!!! So, is that function, when
is this first derivative equal to zero. Well, this first derivative is a rational function
and when your rational function, when you have a fraction, a fraction is equal to zero
when the numerator is equal to zero. So I am going to carry this up and I am going to
set 3x plus 18 equal to zero. That means that 3x is equal to negative 18 and x is equal
to negative six. Well that is the value, that is the x coordinate or c where our first derivative
is equal to zero. Now, just to make sure that I have not done something crazy here, I was
looking for a c value which is just an x value that is between zero and negative nine. Good!
And of course that this function is only going to be defined for values that are greater
than or equal to negative nine. So you want to make sure that this answer that you get
fits within that interval of (a,b) that we were looking for. Also make sure of course
that it is within the domain of the original function. Maybe your textbook is going to
throw some kind of crazy tricky problem in there to see if you are checking those answers
when you are done. One more example. Before we get to the next example, I was erasing
the board and I thought if you went ahead did that problem or wanted to do that problem
with the product rule you might want to see what that derivative process looks like. When
I got to my first derivative, I did not need to simplify that radical in the denominator
and have it cancel with any x's in the numerator. Just to give you another look (the correct
one) at finding the derivative of x times the square root of x plus nine. Hopefully
by now you are not still struggling with finding derivatives, but if you do have...just as
another reminder...if you do have two terms in your function and those two terms have
a common factor and that factor maybe has some negative exponents or like we have rational
exponents and negative, ones negative one-half and one is positive one-half, make sure that
do try and factor out the lowest exponent. That may even be the negative exponent. Just
a little tip there for you to make sure that you do not make any careless mistakes in your
work or help you maybe understand some more of your homework problems. Here now finally
is our last example. Bam! Last example. Determine whether Rolle's Theorem can be applied to
function f on the closed interval [a,b], if so, find all values of c in the open interval
of (a,b) such that the derivative of f of c is equal to zero. So our function here is
f(x) is equal to cosecant of x. Our closed interval is [pi/2,5pi/2]. If you really know
your trig functions, you will know to just basically say that I cannot apply the Rolle's
Theorem, explain an move one. But, we are going to walk through a couple steps just
to, you know, refreshen some of your trig skills if you need it and ultimately get to
the point of why doesn't this work. Well, here is our value of a and b, our low and
upper limit of our closed interval. We are going to take the cosecant, try and find the
cosecant of pi/2. Well, if you don't know what that answer is you might want to convert
cosecant into sine which you should remember off of your unit circle. This is one over
the sine of pi/2. The sine of pi over two is equal to one. Remember your radians. This
is zero radians, this is pi over four radians, pi over two radians or ninety degrees. The
ratio, the sine ratio is y over r so your y value is one, your radius is one on the
unit circle, and thus your answer is one. Good! So now let's check 5pi/2. The cosecant
of 5pi/2 is again over the sine of 5pi/2. I write a lot better, not trying to do this
backwards. There we go. The sine of 5pi/2. That is zero radians, pi over two, 2pi over
two, 3pi over two...there the sine would be negative one, 4p over two, and then finally
5pi over two. So, we are back up at ninety degrees, well, the positive y axis. The sine
of 5pi/2 again is equal to one. So f(a) is equal to f(b). The function at pi over two
is the same as 5pi over two. So, why can't we apply the Rolle's Theorem? Well, I just
used the sine function to find answers for cosecant. That is because sine and cosecant
are reciprocals and there graphs are therefore very closely related. Your generic sine graph,
the sine of zero is zero. You go to pi/2, you get a sine value of one. Then you go to
pi and it comes back down to zero. 3p/2 you are at negative one, as I just indicated with
my little hand motion there as I was going through the problem:) At 2pi you are back
to a value of zero, so this is the first period of sine. Then, 1/2, 2/2, 3/2, 4/2, 5pi over
two you are back to a value of one. Well, there are our values of one that we found
for the cosecant at pi over two and the cosecant of 5pi over two. But, I am not looking at
the cosecant graph yet, I am looking at the sine graph which helps guide our graph for
cosecant. Every where that the sine function passes through its middle point term, not
the max or the min values, but the middle values, in this case those are the x intercepts
because there is no vertical shift, you get these vertical asymptotes. So, this is actually
the graph of cosecant. I ran out of room there. As you can see, that cosecant function is
not continuous on the closed interval from pi/2 to 5pi/2 and that is why the Rolle's
Theorem cannot be applied even though f(a) was equal to f(b). So, even though I can find
a place here at 3pi over two where the derivative would be equal to zero, it is not Rolle's
Theorem (Rolle's didn't guarantee it) because we do not have a continuous function. So,
I am Mr. Tarrou. Go Do Your Homework!!!