Welcome to a lesson on Rolle's Theorem.
Let's go ahead and take a look
at what Rolle's Theorem states.
Let the function f be
continuous on the closed
interval from a to b,
and differentiable on
the open interval from a to b.
If f of a equals f of b, then there is
at least one number c in the open interval
from a to b such that f
prime of c is equal to zero.
Let's take a look at this sketch.
If f of a is equal to f of b,
you can see the secant line
through those two points
would be horizontal and therefore would
have a slope of zero.
And what this theorem
states is if that's true,
and the function is
continuous and differentiable
on that interval, there must be some other
value c where the tangent line
would also have a slope of 0,
meaning the derivative of the function
would be zero at c
meaning the derivative
at c would equal zero.
So regardless of what
this function looks like,
if it's continuous and
differentiable, there
will be at least one value c where the
derivative would equal zero.
If I was to sketch a different function,
maybe something like this,
there would be more than one value of c
because we have a horizontal
tangent line here,
here, here, and here,
so there would be four values
of c rather than just one.
So hopefully when you take
a look at this sketch,
this idea makes sense that
this would have to be true
where there's at least one value
of c where the derivative
would equal zero.
Let's go ahead and give it a try.
So as you determine whether
Rolle's Theorem can be applied,
if it can then find all values of c where
the derivative is equal to zero.
So the first step is
to determine if f of 1
is equal to f of 2.
If these aren't equal we
cannot apply Rolle's Theorem.
Well this equals zero.
And f of two, four minus
six plus two is also
equal to zero.
So f of one equals f of two.
This is a continuous
differentiable function on
this interval so we can
apply Rolle's Theorem.
So what we're going to
do is find the derivative
and set it equal to zero.
The derivative would be two x minus three.
And we wouldn't know when
this would equal zero.
So we solve for x, we
add three to both sides.
Divide by two.
So we have x equals three halves,
so c is equal to three halves meaning
at x equals three halves we should have a
horizontal tangent line.
Let's go ahead and take a look.
So here's the interval
that we're considering,
from one to two.
F of one equals f of two, it's continuous
and differentiable and
you can also see that
x equals 1.5, we have a
horizontal tangent line,
therefore the derivative would equal zero.
Let's go ahead and try another one.
So first step is to determine if f of zero
equals f of two.
So we'd have zero squared
minus two times zero
times e to the zero, that's
zero times one of zero.
F of two would equal two
squared minus two times two.
Well that's going to be zero.
Zero times e squared is also zero.
Next, this graph is
differentiable and continuous
on this interval.
So now we can apply Rolle's theorem.
So we'll find the derivative and see where
it equals zero in this interval.
We will have to apply
the product rule here.
This is our first function,
this is our second function.
So f prime of x equals the first function,
times the derivative of e to
the x which is e to the x,
plus the second function
times the derivative
of x squared minus two x,
that would be two x minus two.
Before we set this equal
to zero we can factor this.
There's a common factor of e to the x.
We'd be left with x squared
minus two x plus two x
minus two equals zero.
Combining our like terms here,
we'd have e to the x times x
squared minus two equals zero.
Well e to the x is never
going to equal zero.
So we need to set x squared
minus two equal to zero.
Let's go ahead and do that up here.
So we'll isolate the x squared,
take the square root of both sides.
X equals plus or minus
the square root of two.
But negative square root
two is not in this interval,
so our value for c would be
positive square root two.
Let's go ahead and check this graphically.
Again we're considering this function on
the interval from zero to two.
F of a equals f of b, that's correct.
And the square root of
two is right about here
where we would have a
horizontal tangent line,
meaning that a derivative
would equal zero.
Okay let's try one more.
Here we have a trig function.
So let's first determine
if f of pi over six
equals f of pi over three.
So it's going to be sin
two times pi over six
and sin of two pi over three.
Well the sign of pi over three is going
to be square root three over two.
And the sin of two pi over three is also
square root three over two.
These both have a reference
angle of 60 degrees
in the first and second quadrant.
Remember that sin is positive in both
the first and second quadrant.
Next, this function is
continuous and differentiable
on this interval, so we need to determine
where the derivative would equal zero.
This is a composite function.
F prime of x is going
to equal the derivative
of sin two x, would be cosin two x,
times the derivative
of the inner function,
which would give us two.
We want to know when
this is equal to zero,
so we have two cosin two x equals zero.
Divide both sides by two.
When is cosin two x equal to zero?
Remember, we're only talking about
x values on this interval.
And we know that the
cosin function is equal
to zero at pi over two radians.
So since we know our
angle, it must be equal
to pi over two radians.
What we'll do is we'll set two x equal
to pi over two radians and solve.
So multiply both sides by one half.
X would be equal to pi over four radians.
So the value of c is pi over four radians.
That's where the derivative
would equal zero.
Then again, let's check this.
We're only considering this
function from pi over six
to pi over three.
F of a equals f of b.
We can see if it's continuous
and differentiable.
And we can also see right
here at pi over four
we would have a horizontal tangent line,
and therefore the
derivative would equal zero.
The next video will be on
the mean value theorem,
which is just an extension
of Rolle's Theorem.
