Video Transcript
In this video, we’ll learn how to
interpret and use the mean value theorem and Rolle’s theorem. We’ll begin by looking at how
Rolle’s theorem works. And how it leads us into the mean
value theorem, before considering a number of examples of the application of this
theorem. Rolle’s theorem was first published
in 1691 by French mathematician Michel Rolle. He was a vocal critic of calculus,
calling it inaccurate and a collection of ingenious fallacies, though he did
eventually change his opinion.
Rolle’s theorem says that if 𝑓 is
a function that satisfies the following three hypotheses. That is, it’s continuous on the
closed interval 𝑎 to 𝑏 and differentiable on the open interval 𝑎 to 𝑏. And 𝑓 of 𝑎 is equal to 𝑓 of
𝑏. Then, there exists a number 𝑐 in
the open interval 𝑎 to 𝑏, such that the derivative of 𝑓 evaluated at 𝑐, that’s
𝑓 prime of 𝑐, is equal to zero. In other words, if the function
satisfies these criteria, then there’s a point on the graph in this closed interval
for which the slope of the curve is zero. The tangent of the curve at this
point is horizontal. It also has a clear physical
meaning. If a body moves along a straight
line and after a certain period it returns to the starting point. Then there’s a moment in this
period of time where the instantaneous velocity of the body must be equal to
zero.
Now, this theorem is seldom used as
it tells us of the existence of a solution, but not how to get there. It is, however, extremely useful in
helping us to derive the mean value theorem. Let’s have a look at that. This theorem was first stated by
another French mathematician, Joseph-Louis Lagrange. Though we don’t know whether he was
as critical of calculus as Michel Rolle. It says that if 𝑓 of 𝑥 is a
function which is continuous on some closed interval 𝑎 to 𝑏 and differentiable at
every point of some open interval 𝑎 to 𝑏. Then there is a point 𝑐 in this
open interval. Such that the derivative of 𝑓
evaluated at 𝑐, that’s 𝑓 prime of 𝑐, is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏
minus 𝑎. Notice that this right-hand side is
essentially the slope formula. 𝑚 equals 𝑦 two minus 𝑦 one over
𝑥 two minus 𝑥 one, written using function notation. And what this formula therefore
tells us is that there’s some 𝑥-value in the open interval 𝑎 to 𝑏, where the
slope of the tangent line is the same as the slope of the secant line passing
through the two end points of the closed interval. Let’s prove this.
We’ll let 𝑔 of 𝑥 be the secant
line to 𝑓 of 𝑥 passing through the two end points of our closed interval at 𝑎, 𝑓
of 𝑎 and 𝑏, 𝑓 of 𝑏. We can look to find the equation
for 𝑔 of 𝑥 using the formula for the equation of a straight line. That’s 𝑦 minus 𝑦 one equals 𝑚
times 𝑥 minus 𝑥 one. We can let 𝑥 one be equal to
𝑎. And thus 𝑦 one is equal to 𝑓 of
𝑎. We can also change 𝑦 to 𝑔 of
𝑥. And we know that the slope 𝑚 is
given by the change in 𝑦 over the change in 𝑥. That’s 𝑓 of 𝑏 minus 𝑓 of 𝑎 over
𝑏 minus 𝑎. And we obtain that 𝑔 of 𝑥 minus
𝑓 of 𝑎 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎 times 𝑥 minus 𝑎. And we can solve for 𝑔 of 𝑥 by
adding 𝑓 of 𝑎 to both sides.
Next, we introduce a new function
ℎ. And this is defined as the vertical
distance between 𝑓 of 𝑥 and the secant line 𝑔 of 𝑥. So ℎ of 𝑥 is defined as 𝑓 of 𝑥
minus 𝑔 of 𝑥. Our next job is to substitute the
expression for 𝑔 of 𝑥 into this equation for ℎ of 𝑥. But how is this useful? Well, notice that ℎ of 𝑎 and ℎ of
𝑏 are equal. They’re both zero since the
vertical distance between the function 𝑓 of 𝑥 and the secant line is zero at these
end points. And now we should be reminded of
Rolle’s theorem. ℎ of 𝑥 is continuous and
differentiable on the open interval 𝑎 to 𝑏. And so there must exist some value
of 𝑥, 𝑐 in this interval, such that ℎ prime of 𝑐, the derivative of ℎ evaluated
at 𝑐, is equal to zero. So let’s differentiate both sides
of this equation.
The derivative of ℎ is ℎ prime. And the derivative of 𝑓 is 𝑓
prime. Now let’s look carefully at
everything inside this bracket. When we distribute these
parentheses, we end up with this quotient, which is a constant being multiplied by
𝑥 and then being multiplied by another constant. Similarly, 𝑓 of 𝑎 is also a
constant. We know that the derivative of a
constant is zero. And we also know that the
derivative of some constant multiplied by 𝑥 is simply that constant. So we see that ℎ prime of 𝑥 is
equal to 𝑓 prime of 𝑥 minus 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. From Rolle’s theorem, we can say
that ℎ prime of 𝑐 must be equal to 𝑓 prime of 𝑐 minus 𝑓 of 𝑏 minus 𝑓 of 𝑎
over 𝑏 minus 𝑎, which must be equal to zero. And we’ve actually proved the mean
value theorem. If we solve for 𝑓 prime of 𝑐, we
obtain that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
Now that we’ve proved the mean
value theorem, let’s have a look at an example of how to establish where the theorem
actually applies.
Does the mean value theorem apply
for the function 𝑦 equals two 𝑥 cubed minus four 𝑥 plus seven over the closed
interval zero to five?
To use the mean value theorem, two
things must be true about our function 𝑓 of 𝑥. It must be continuous over the
closed interval 𝑎 to 𝑏. And it must be differentiable over
the open interval 𝑎 to 𝑏. Well, the function two 𝑥 cubed
minus four 𝑥 plus seven is indeed continuous over the closed interval zero to
five. It’s a simple cubic graph that
looks a little like this over our closed interval. And to check for the second
condition, we’ll see what happens when we do differentiate with respect to 𝑥. The derivative of two 𝑥 cubed is
three times two 𝑥 squared. That’s six 𝑥 squared. And the derivative of negative four
𝑥 is negative four. So we obtain d𝑦 by d𝑥 to be equal
to six 𝑥 squared minus four. This is indeed defined over the
open interval zero to five. And we can say yes, the mean value
theorem does indeed apply.
This example demonstrates that we
simply need to check the conditions required for the mean value theorem to establish
whether we can use it for a given function. Now let’s look at an example of how
we might use it.
For the function 𝑓 of 𝑥 equals 𝑥
cubed minus four 𝑥, find all the possible values of 𝑐 that satisfy the mean value
theorem over the closed interval negative two to two.
Remember, the mean value theorem
says that if 𝑓 is a function which is continuous over some closed interval 𝑎 to 𝑏
and differentiable at every point of some open interval 𝑎 to 𝑏. Then there exists a point 𝑐 in
that open interval such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over
𝑏 minus 𝑎. Our 𝑓 of 𝑥 is equal to 𝑥 cubed
minus four 𝑥. And our closed interval is from
negative two to two. We’re looking to find the value of
𝑐 such that the derivative of our function evaluated at 𝑐 is equal to 𝑓 of 𝑏
minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. That’s 𝑓 of two minus 𝑓 of
negative two over two minus negative two. So we’ll do two things. We’ll evaluate this quotient. And we’ll also find an expression
for the derivative of our function. 𝑓 of two is two cubed minus four
times two. And 𝑓 of negative two is negative
two cubed minus four times negative two. And actually, this gives us a value
of zero. So in order to find the values for
𝑐 such that 𝑓 prime of 𝑐 is equal to zero, let’s find the derivative of our
function.
The derivative of 𝑥 cubed is three
𝑥 squared. And the derivative of negative four
𝑥 is negative four. So 𝑓 prime of 𝑥 is three 𝑥
squared minus four. And we can say that 𝑓 prime of 𝑐
is equal to three 𝑐 squared minus four. So let’s set this equal to zero and
solve for 𝑐. We begin by adding four to both
sides of our equation to obtain that three 𝑐 squared is equal to four. Next, we divide by three. And we see that 𝑐 squared is equal
to four-thirds. And finally, we take the square
root of both sides, remembering to take by the positive and negative square root of
four-thirds. To obtain that 𝑐 is equal to plus
or minus the square root of four over three. And which we can say is equal to
two over root three and negative two over root three. Notice also that these values of 𝑐
are indeed in the closed interval negative two to two as required by the mean value
theorem.
For the function 𝑓 of 𝑥 equals 𝑥
minus one to the eighth power, find all the possible values of 𝑐 that satisfy the
mean value theorem over the closed interval zero to two.
Remember, the mean value theorem
says that if 𝑓 is a function which is continuous over the closed interval 𝑎 to 𝑏
and differentiable at every point of the open interval 𝑎 to 𝑏. Then there’s a point 𝑐 in that
open interval, such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏
minus 𝑎. Our 𝑓 of 𝑥 is 𝑥 minus one to the
eighth power. And our closed interval runs from
zero to two. We’re looking to find the value of
𝑐 such that the derivative of our function evaluated at 𝑐 is equal to 𝑓 of 𝑏
minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. That’s 𝑓 of two minus 𝑓 of zero
over two minus zero. So we’ll do two things. We’ll evaluate this quotient. And we’ll also find an expression
for the derivative of our function and evaluate it at 𝑐. 𝑓 of two minus 𝑓 of zero is two
minus one to the eighth power minus zero minus one to the eighth power. And we obtain that 𝑓 of two minus
𝑓 of zero over two minus zero is zero divided by two, which is just zero.
Our next job is to find the
derivative of our function. And we’ll use the general power
rule. This says that if 𝑔 of 𝑥 is a
differentiable function and 𝑛 is 𝑎 constant real number, such that 𝑓 of 𝑥 is
equal to 𝑔 of 𝑥 to the power of 𝑛. Then the derivative of 𝑓 of 𝑥, 𝑓
prime of 𝑥, is equal to 𝑛 times 𝑔 of 𝑥 to the power of 𝑛 minus one times the
derivative of 𝑔 of 𝑥. That’s 𝑔 prime of 𝑥. The derivative of our function 𝑥
minus one to the eighth power is therefore eight times 𝑥 minus one to the seventh
power multiplied by the derivative of 𝑥 minus one, which is one. And so we see that 𝑓 prime of 𝑥
is eight times 𝑥 minus one to the seventh power.
We see that 𝑓 prime of 𝑐 is eight
times 𝑐 minus one to the seventh power. And remember, we found that 𝑓 of
𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎 is zero. So we’re going to set this equal to
zero and then solve for 𝑐. We divide by eight. And we see that 𝑐 minus one to the
seventh power is zero. And we can take the seventh root of
both sides to see that 𝑐 minus one is equal to zero, which means that 𝑐 equals one
is the only value of 𝑐 that satisfies the mean value theorem. Notice it falls in the closed
interval zero to two, as required.
In our final example, we’ll look at
how this theorem can be applied to a contextual question.
A rock is dropped from a height of
81 feet. Its position 𝑡 seconds after it is
dropped until it hits the ground is given by the function 𝑠 of 𝑡 equals negative
16𝑡 squared plus 81. Determine how long it will take for
the rock to hit the ground. Find the average velocity of the
rock from the point of release until it hits the ground. And find the time 𝑡 according to
the mean value theorem when the instantaneous velocity of the rock is equal to the
average velocity.
The rock will reach the ground when
its position 𝑠 of 𝑡 is equal to zero. We can therefore set this
expression negative 16𝑡 squared plus 81 equal to zero and solve for 𝑡. We add 16𝑡 squared to both sides
and then divide through by 16. And we obtain 𝑡 squared to be
equal to 81 over 16. We then take the square root of
both sides, remembering to take by the positive and negative square root of 81 over
16. And we see that 𝑡 is equal to plus
or minus nine-quarters. Now, we can actually disregard
negative nine-quarters since we’re working in time. And we find that the rock hits the
ground after nine-quarters of a second.
Our next job is to find the average
velocity of the rock over this period of time. The definition for average velocity
is total displacement divided by time taken. The displacement of our rock is its
change in position. That’s negative 81 feet. And it takes nine-quarters of a
second to travel this far. So the velocity is negative 81
divided by nine over four. Remember, to divide by a fraction,
we can multiply by the reciprocal of that fraction. So we have negative 81 times four
over nine. And then we cancel this factor of
nine. And so we obtain that the average
velocity of our rock is negative 36 feet per second.
For the final part of this
question, we’ll need to quote the mean value theorem. Remember, this says that if 𝑓 is a
continuous function over some closed interval 𝑎 to 𝑏 and differentiable at every
point of that open interval 𝑎 to 𝑏. Then there’s a point 𝑐 in this
interval, such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus
𝑎. We know that the average velocity
is negative 36 feet per second. That’s the equivalent of this
quotient. The instantaneous velocity can be
found by differentiating our function for position. That’s 𝑠 prime of 𝑡 equals
negative 32𝑡. In this case then, we can say that
𝑠 prime of 𝑐 is equal to negative 32𝑐. And we obtain the equation negative
32𝑐 equals negative 36. We solve for 𝑐 by dividing both
sides by negative 32. And we find that the time at which
the instantaneous velocity of the rock is equal to the average velocity is equal to
nine-eighths of a second.
In this video, we briefly discussed
Rolle’s theorem. And we said that if a function 𝑓
satisfies three criteria. That is, it’s continuous over the
closed interval 𝑎 to 𝑏, differentiable on the open interval 𝑎 to 𝑏, and 𝑓 of 𝑎
equals 𝑓 of 𝑏. Then, there exists a number 𝑐 in
that open interval such that the derivative of 𝑓 evaluated at 𝑐 is equal to
zero. We also saw that we can use Rolle’s
theorem to prove the mean value theorem. And this says that if 𝑓 of 𝑥 is
continuous over a closed interval 𝑎 to 𝑏 and differentiable over an open interval
𝑎 to 𝑏. Then there exists some number 𝑐 in
that open interval such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over
𝑏 minus 𝑎.
