Rolle's Theorem

Uploaded by TheIntegralCALC on 27.06.2011

Hi, everyone! Welcome back to Today we’re going to be doing a Rolle’s
theorem problem. And in this one, we’ve been asked to prove that the function f(x)
= x^2 – 2x satisfies the conditions of Rolle’s theorem on the range 0 to 2.
First of all, let’s talk about what it means to satisfy Rolle’s theorem. Rolle’s theorem
has three premises and then a conclusion so it says that if three things are true, then
a fourth thing must be true. So let’s go ahead and talk about what that is. Rolle’s
theorem tells that if a function is continuous over a certain range, in this case our range
is 0 to 2, if it is differentiable meaning that  you can differentiate it on the entire range 0 to 2, it has a derivative
and three, if f(a) = 0 = f(b). If these three things are true, then it means that at some
point c on the range a to b, it means that the derivative at the point c is going to
be equal to zero. Let’s talk about what this means and we can pull up the graph of
this particular function. So what this means is that in order for this
function to satisfy Rolle’s theorem on the range zero to 2, the function is continuous
over the range 0 to 2. So we’ve got zero here and 2. So the function is continuous
over this entire range. And we can see pretty much just by looking at it that it is continuous.
Remember that continuity means that there’s no breaks in the graph, there’s no holes
in it. So this function is continuous on the range 0 to 2; that it’s differentiable which
in this case the function is because there’s no vertical asymptote and there’s no break
or point like an absolute value or something like that. It’s differentiable. This third
condition here that f(a) = 0 = f(b) means that at the endpoints 0 and 2, the function
is equal to zero;, meaning that it’s touching the x axis at that point; that it’s equal
to zero. So the endpoints of this range have to be right on the x axis. So that’s what
that third condition means. And what Rolle’s theorem is saying is if all three of those
things are true, then there’s got to be some point c within the range between the endpoints that the derivative
of the function is going to be equal to zero which means that the tangent line at that
point is going to be horizontal. So that’s all that that means. We’ve got c here. The
point c corresponds to this point here in the graph, 1. And if we draw the tangent line
through that point, we can see that the tangent line is horizontal.
So basically, just to recap, all Rolle’s theorem is saying is that if you’ve got
a range a to b, if the function is continuous and if it’s differentiable and if the endpoints
are on the x axis, then there’s going to be some point inside the range where the tangent
line of the function is horizontal or the slope of the tangent line is equal to zero.
This is what this is saying here; that the slope of the tangent line is equal to zero.
So in order to prove that this function satisfies this theorem, here’s what we’ll do. We
will take the derivative of f(x). So that’s how you prove it. We take the derivative of
f(x) and of course when we do that you get 2x – 2 and then you’re going to set that
equal to zero and solve for x. So we say 0 = 2x – 2. We add 2 to both sides to get
2 = 2x and then we divide both sides by 2 and we get 1 = x. So that’s the point c
in the range a to b where the tangent line is horizontal. We know that that’s the point
but we need to prove that it’s continuous. Remember that continuity is going to exist
unless there is a point in the graph of which it’s undefined. So if your function is a
fraction, it would be the point at which the denominator would be equal to zero that the
function would be discontinuous. If you’ve got a vertical asymptote, if you have a square
root sign and if there are negative values within the square root sign, then you might
have discontinuity in your function. In this case, none of those things exist for us so
we know that the function is continuous. We also know it’s differentiable and we know
that there are no holes, there are no vertical asymptotes; there’s no point in the graph
at which it breaks. You can take your pencil and draw in one continuous motion without
picking your pencil up. So we know that it’s also differentiable. And we know that f(a)
= 0 = f(b) and we can prove that by plugging in the endpoints of our range zero and 2.
So if we plug in 0 and 2 to the original function, we’d get f(0) = 0^2 – 2(0). So that is
equal to zero. And if we plug in 2, we’d get 4 – 4 which is also going to be equal
to zero. So we’ve now proved that both of the endpoints a and b are equal to zero. So
that condition is satisfied. So we know all three conditions are satisfied, which proves
the existence of a point c. And in order to find that point, we just take the derivative,
set it equal to zero and solve it for x and we know that there is a point x = 1 where
the tangent line is horizontal and again, if we look at our graph, we can see that that
point is right here at x = 1. The tangent line at that point of the graph is horizontal.
So that’s it. I hope this video helped you guys and I will see you in the next one. Bye!