How Do You Use the Mean Value Theorem? (and Rolle’s Theorem)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[Music] so [Music] the mean value theorem and its special case rolls theorem
are crucial theorems in the calculus they provide a mechanism for solutions to
numerous applications and a resource that helps us prove countless other
celebrated theorems in analysis understanding these theorems
is the topic of this video hi everyone welcome back uh this is the calculus one
explore discover learn series this episode
is mean value theorem with rose theorem and examples

00:01
um so i’m really excited about today’s um today’s talk today we’re going to talk
about rolls theorem first and then we’re going to follow that up
with mean value theorem and then i’m going to make sure that we
understand the difference between the two and how they’re related and then we’re
going to talk about uh applications of the mean value theorem
in terms of mathematics and you know the real world and how you
could see the mean value theorem and then stick around to the end where
we’re going to go over some exercises so let’s go ahead and get started
okay so up first i want to mention that the
extreme value theorem is an important theorem that i’m going to talk about
right now a little bit uh or i’m going to use a little bit when
i talk about rolls theorem so the extreme value theorem was in a

00:02
previous episode um so that episode is very important for this episode
so if you haven’t seen that well um in the description below i have the
playlist link and you should be able to see a video
that says extreme value theorem so make sure to watch that episode first
i talk about also the differentiation episode i think that’s episode three
and so yeah we need those um we need to know that material before we
really take a hard look at this so if you haven’t done so check it out
if you have let’s go here we go so the extreme value theorem guarantees
the existence of a maximum minimum minimum value
of a continuous function on a closed interval
and the next theorem rolls theorem uh it’s going to guarantee the existence of
an extreme value theorem on the interior of a closed interval and

00:03
we’re going to have two other con conditions besides continuity
so role serum is going to say that if a function is also differentiable
on an open interval it’s continuous at the endpoints
and if the function values are equal at the end points
so all of that combined together all three assumptions
then it has to have at least one horizontal tangent and so we’ll see that now
a special case of all this is of course that the function is constant
the for if the function is just a horizontal line
then of course we’re gonna have the derivative is zero or horizontal tangent so
basically the point is that rolle’s theorem guarantees us the existence of
at least one point in the interval where there will be a horizontal tangent line
so we’ll see in this video upcoming that rule serum is the special case of
the mean value theorem for when the values of the function are

00:04
the same at the endpoints of the interval so what the heck does all that mean
well let’s get started and see what one through five let’s see what it all means
so here we go here’s here’s the beginning let f be a function that’s
continuous on the closed interval now just to refresh your memory on what
that means is means that the function f is continuous
for every number in the open interval a b and then it’s continuous um you know
from the right at a and from the left and b and we’re also assuming it’s a
differentiable on the open interval a b [Music]
um and then the third condition that we’re assuming
is that the height of the endpoints are the same in other words f of a
equals f of b so the conclusion of rolle’s theorem says
if if those three assumptions hold then there exists

00:05
a real number c between a and b where the derivative is zero so
i’ll sometimes say the derivative is zero in other words
we have a horizontal tangent line okay so here’s the understanding of the proof
first case is as we mentioned on the on the first page
if f is a constant then in fact the derivative is zero
because the derivative of a constant is zero so roll’s theorem doesn’t say
anything special for constant functions this just states
a true fact but it’s not very enlightening
but if your f of x is greater than f of a for some x and a b then by the extreme
value theorem so this is what i’m saying it’s important that you
fully understand that that episode if so by the extreme value theorem we
have to have a maximum value somewhere and that’s exactly what happens because

00:06
f of x is greater than f of a so we have to have f prime c equal to zero
and the reason why is because of our second assumption the derivative exists
everywhere so knowing that the derivative exists anywhere
everywhere and that interval we know we don’t have any sharp corners
so we looked at that in close detail in the last video
similarly if f of x is less than a well we still apply the extreme value
theorem and we know that f has to obtain its absolute minimum value
somewhere in the interval and this is where the slope of the tangent line
is zero so this is the understanding that rolls theorem
um to a great extent follows from extreme value theorem
okay so let’s verify rules theorem for this function right here so when i’m
looking at this function right here i have a polynomial it’s a third degree

00:07
polynomial but it has a restricted domain it’s a closed interval
so to to verify role’s theorem for this we have to do three things we have to
check f is continuous on minus two to two f is differentiable [Music]
on the open interval [Music] and then the last thing to check is that
f of minus two is equal to f of two and so what are those values so this
will be four minus four times minus two minus minus four squared [Music]
plus minus two to the third so we have to figure out what that
number is and the same thing we have to plug in

00:08
2 and find out what that number is so when we calculate this up we’re going
to get 0 and when we substitute in two we’re going to get four minus four
times two minus um [Music] four squared plus four to the third
and in this case we also get zero so we’ve checked that f of a
where a is minus two and f of b where b is two
the endpoints both agree they’re both equal to each other so f of minus 2
equals f of 2. so we’ve checked this and this and this so we checked the
three things that we needed to check for rolle’s theorem so now we know that

00:09
rolle’s theorem holds and so now we can say therefore there exists
so there exists so by rose theorem there exists a c somewhere between -2
there exist c in minus two to two such that the derivative is
zero the derivative at c is zero so you know what is the derivative
the derivative is so we know that c exists by rolle’s theorem
but actually because we’re given a very particular function we can go
we can go try to find it if we want so this derivative will be minus four
minus 2x and then plus 3x squared so where is the derivative zero this is

00:10
a quadratic equation now quadratic equations either have two real solutions
one real solution or no real solutions but we know
that by rolls theorem there is at least one solution to this
in this interval and so we can go and try to solve that
so we can just use you know quadratic formula whatever and we’ll be able to find
the solutions to that so you know x is equal to um the negative negative 2
plus or minus square root of minus 2 squared minus 4
times the a times the c minus four all over two times the three and so yeah

00:11
using the quadratic formula that’s uh minus 2 there and that is -2 also and so
using the quadratic formula there we can go and so sometimes this will not be
positive and so you might get no solutions to this
because we might have square root of a negative there in any case
rose theorem says we have to have at least one c
even if you have a hard time finding it in fact
imagine for a moment if our original function
wasn’t x to the third what if it was x to the sixth or something like that
then if you were to go take the derivative
you would get a derivative of one degree less
and then it could be very difficult to solve because this is the quadratic
we have a formula but this could be it could be very difficult to solve
the derivative equal to zero and to find that
actual x or that actual c where the derivative

00:12
is zero so the rose theorem is very powerful because
it tells you something exists even if you may not be able to actually go and
find it to find the actual value for it just knowing that something exists
can be extremely powerful all right so let’s write this up
so first thing i like to take a peek at the graph
so i’m looking at the graph of f 4 minus four x minus x square plus x to the
third and that’s on minus two to two so as you can see it goes up and then it
comes back down and it goes back up again
and you can see that the height at minus two
and two is 0 and that’s important for rolls theorem
it’s continuous function it’s differentiable and the
endpoints agree so somewhere between -2 and 2 there has to be a number
where there is a horizontal tangent line in

00:13
actually c2 one for the max and one for the min
so here notice that f is continuous on the closed interval
and f is differentiable on the open interval and that when we plug in minus two
and we plug in two we get out zero so we’ve checked all three conditions to
use rolle’s theorem and so we know there is at least one c in that open interval
where the derivative is zero and so we go take the derivative
and we set it equal to zero and then we solve and we actually
simplify that and we actually find those two c’s
now sometimes you’ll go and solve that quadratic
and you may get two answers you may get one you may get none
but don’t forget to always go back and look at your original domain
and know that both of them or or whatever you claim

00:14
is on your domain all right so let’s look at a different type of example now
we’re going to use rolls theorem to show that
this equation right here has only one real root we’re going to show that it
has one and it only has one so as i was mentioning before if your
polynomial is degree five or six or seven or so on
it they can be very difficult to solve to find their roots
and so you’ll wanna have ways of knowing if something exists
or not that’s extremely important for example
suppose you’re unable to find the exact value and instead you want to rely upon
approximation methods and you’re fine with approximating a
root say to five or six decimal places that that may be exactly what you want
but before you go apply an algorithm you may need to know that something
exists and how many of them exist so using these theorems can be very
important before you go apply an algorithm which may actually

00:15
approximate the root okay so let’s see how we can do this
so we’re going to use intermediate value theorem
which was covered in the episode two on continuous functions and we’re going to
use rose theorem which we just talked about here so this is going to be
another example of using roles theorem but we’re also going to use intermediate
value theorem now i wanted to put them side by side so
you can see the difference and kind of kind of review
what those theorems say so f says uh intermediate value theorem says if f is
continuous on the closed interval in fact they both start that way don’t they
let f be continuous on a closed interval a b um and then for the immediate value
theorem l is some number between f of a and f
b so think of f of a and f of b on the y
axis right those are some heights and so l is some height in between those two

00:16
and then it says there’s some c where f c equals l and that’s all based upon
continuity if you can draw it without picking up your pencil
you’re going to find that c where you get that height l
and then roll’s theorem not only do we have continuity
but we also are we also need differentiability on the open
and the endpoints are equal and then there’s one place
where at least one place where the derivative is zero
okay so let’s see how to use these two theorems to come up with the fact that
this equation right here has exactly one real root and so
again the idea might be to go and try to approximate the root
but knowing that it only has one real root is is is a very important step
okay so show that this equation has exactly one real root
so i’m going to take a peek at the graph now the graph isn’t really used in this

00:17
proof here but i think understanding this example you can have
that visualization in mind so that you can kind of understand
what’s happening so there’s the actual graph
now when we look at this graph right here
i’m looking at this graph and i’m asking myself a question
right because the graph looks like this [Music]
and i’m asking myself you know because it says it has exactly one real root
is sitting right here and here’s one and here’s minus one over here and
there’s the graph now but i’m asking myself how do i know that’s the graph
i mean how do i know it doesn’t come back down over here somewhere else
how do i know that there isn’t some other route we
our our goal is to show it only has one in fact how do we also know that this
one doesn’t come up and there isn’t another route over here i mean how do we

00:18
know that this is the only route so that’s the question how do we know
there isn’t some other route [Music] how do we know there’s even one route
so those are the type of questions that we have to answer first of all how do we
know that there’s one and then the second one
is well how do you know that there’s only one
okay so when i say that’s the graph there
we’re not entirely sure that that is the actual graph
that’s just kind of a visualization to get us started so we
we can see give more life to the words i’m about to say
but the words are the argument okay so the first thing is i notice
is that this is a polynomial okay so x to the third plus x
minus one that is the polynomial and we have some
very powerful theorems for polynomials we know polynomials are continuous
everywhere and we know that polynomials are differentiable
everywhere we’re all real numbers so if i start looking at this polynomial

00:19
on and restricting it over different domains
i know it’ll be continuous and i know to be differentiable
so that means that we have these two theorems which will apply the
intermediate value theorem and the roles theorem they both apply so
it’s very important that we communicate that with our reader and for
yourself it’s very important that you check that
and that you understand that first before you can go use those big theorems
we have to actually understand our function those are that
that’s a polynomial we understand polynomials
all right now the next thing is to use intermediate value theorem
remember we needed f of a and f of b and we needed some number l in between
right so i’m gonna check that f of zero and i get out minus one and when i look
at f of one i get one so if we can go back here this is

00:20
minus um this is when we’re zero right here where we’re at minus one
right our function is x to the third plus x minus one so if we plug in zero
we’re out at minus one so we’re right here
so when i look at f of zero right here i get a height i get a minus one and
when i look at f of one right here f of one is one
if we plug in one we get a height of one right here
so here’s my here’s my two heights here and here’s my
l which is zero which is in between those two
so the intermediate value theorem says that this one is negative
this one is positive right that’s negative one and that’s one that’s positive
then this is continuous so somewhere in between it had to go through that zero
so by the intermediate value theorem we have to pass through that zero

00:21
there can be no hole there you know that’s not continuous
so since we know it’s a polynomial there’s no holes in it
it’s continuous it has to hit zero somewhere
okay so that’s the first part there by the intermediate value theorem
there’s some root there’s a c in zero one where the function is zero
so we have at least one root by the intermediate value theorem
okay so there exists at least one one solution
so now we’re going to try to prove that well
wait why not two solutions how do we know that the function
doesn’t come back down like that how do we know that doesn’t happen so
let’s call this first one right here x1 and let’s
think for a moment well maybe it does come back down

00:22
if it does i’ll call that x2 so now i have an interval x1 and x2
and the endpoints of that interval the function is both zero
in other words the function is the same on the endpoints
so again this is a polynomial whatever it looks like whether it comes down or
not it’s a polynomial it’s it’s continuous and it’s differentiable um
on the open interval and it’s continuous at on the closed interval
so rose theorem applies so rose theorem says
that there’s some c in here in between where the slope of the tangent line must
be zero in other words uh i think this function goes up forever
all right but what if it doesn’t what if it comes back down
well if it comes back down it’s going to have to have a horizontal iso
it’s going to have to have a horizontal tangent line
because we know the derivative exists everywhere there’s not there’s not going

00:23
to be a sharp corner there it’s differentiable everywhere so
there’s going to have to be a horizontal tangent line there
well what does rose theorem say i mean uh what is the derivative say so
here’s the function [Music] x to the third plus x minus one row’s
theorem says the derivative has to be zero somewhere
the derivative has to be zero somewhere by rolls theorem but
if you look at the derivative the derivative is three x square plus one
and that is a quadratic that has no real number solutions to it so the
derivative can not never be zero actually so we’re
going to have a problem with this x2 existing
the x2 cannot exist if the x2 exists then we’re going to have to have a place
where the derivative is 0 by rolle’s theorem but if we look at the derivative
it can never be zero so we can’t have both of those things happening

00:24
that the derivative is zero and the derivative is not
zero right there’s no solution to this if you try you get x squared equals
minus one x squared equals minus one third the point is just that
the point is that you’ll never get a perfect squared is equal to a negative
number right there’s no solution to this so if you try to make a solution x2
you’ll run into a problem so let’s write that up
to prove that f of x equals zero for only one
x right we we show there’s at least one already
but to prove that there’s only one we assume that there’s two in x1 and then x2
and we prove that this cannot happen so we’re going to assume that x1 and x2
that there are two is solutions and we’re gonna find a big problem with that
so if there’s two solutions then they’re both zero because that’s what it
means to be a solution and i’m gonna assume that x1 is less than x2

00:25
so now you might say well we didn’t argue over here
well if it comes back up over here then i’ll say
okay this is my x1 and now this is my x2 over here now if it comes back up
then i’m still going to have to have a horizontal tangent line
by rolls theorem but if you still look at the same derivative
it cannot have a solution where it’s equal to zero
so again this cannot come back up so it doesn’t matter if this is x1 and
this is x2 or this is x1 and this is x2 it doesn’t matter in either case we’re
going to get the same problem so i’m going to assume that x1 and x2
are solutions that means that when you plug them in you get 0
and i’m going to assume that x1 is less than x2
so by rolls theorem there exists a c now now when i say by rolls theorem here
what do we got to check we got to check three things

00:26
we check continuity we check differentiability
we check they’re the same on the end points so
rose theorem says there has to be a c somewhere
in the interior where the derivative is zero but if we actually go look at the
derivative we get three x squared plus one and there’s no solution for that
there’s no real solutions for that so in fact that c cannot exist so
there cannot be two so there has to be at most one
there’s exactly one all right let’s take a break [Music]
okay so now we’re going to start talking about the mean value theorem
rose theorem was a warm-up and we’re going to see exactly how
they’re connected with each other hopefully you got the rolls theorem if not

00:27
now would be a good time to maybe rewind it and re-watch it and
and go a little slower um you know change of playback speed or whatever but
let’s go on to the mean value theorem so now we’re going to be given a function
that is differentiable on an open interval and continuous at the end points
so that’s not going to change the mean value theorem is very similar to rolls
theorem we’re still going to have a continuous
on the closed and differentiable on the open interval
the mean value theorem states that there exists a number
in the open interval where the slope of the tangent line
on the graph is the same as the slope of the line through the endpoints
of the on the graph and so there’s a very nice way of saying it um
it looks like a lot of words but actually it’s going to be very intuitive

00:28
so when you see the mean value theorem i think it’s very
important to you know know the three hypothesis know the three hypothesis
um know the conclusion and always have a
an example graph in your mind of what is and what is of what it looks like so
um let’s look at an example graph and see what it looks like and let’s
look at the three hypothesis and the uh the two hypothesis and the
conclusion right so the mean value theorem world’s
theorem had three hypothesis mean value theorem only has two
so for mean value theorem all you need is a continuous function
on a closed interval and it’s differentiable on that open interval
so the conclusion for mean value theorem is there exists a c
so it’s another existence theorem but this time we’re going to say
instead of f prime of c is zero we’re going to say

00:29
f prime at c is equal to the slope of the line
through the endpoints remember that expression right there
represents the slope of the line through the endpoints
or the average rate of change so this is important where
you know what those two things mean on both sides of that equal sign
and so if you’re not familiar with either one or both
might want to check out episode three right
exhaustively cover the left side and the right side
and those rates have changed so sometimes i’ll say the left hand side as
the slope of the secant line uh sorry the slope of the tangent line
and sometimes on the right hand side i’ll talk about that as being the slope
of the secant line through the endpoints all right so the
mean value theorem gives us that c it tells us there exists so just like
the intermediate value theorem just like rolls theorem this is an existence

00:30
theorem all right and so here’s the graphs that you want to kind of have in mind
so i restated the mean value theorem for us let f be a continuous function on a
closed interval and assume it’s differentiable in the open interval
then there exists a c so read that statement over and over again for each
picture for each diagram for the left hand side
we have the graph is in black you can see it’s continuous there’s no holes
there’s no isotopes it’s continuous it’s differentiable
there’s no sharp corners it’s nice and smooth
it just makes a nice pretty arc for the graph in black on the left-hand side
and we can see that there’s a number c right so that equation holds so
the um f prime c that tangent line is in red and you can
see the slope of that tangent line it’s got negative slope it’s going down

00:31
and the slope of the blue line which is the line through the endpoints
of the interval and you can see those are equal to each
other because those are parallel lines because they’re parallel that means the
slopes are equal so we see that the derivative is the the
instantaneous rate of change is equal to the average rate of change that’s
another way of saying it now the mean value theorem gives us the
existence of at least one c but the fact is that there may be lots of c’s
so the diagram on the left shows you where there’s exactly two c’s
so again read through the theorem the graph is in black
and we can see that that curve is continuous on a and b
there’s no holes there’s no isotopes it’s just nice and continuous
and it’s also differentiable there’s no sharp corners
so there exist at least one number c well in this
in this case there’s two there’s a c one and a c two

00:32
where the slope of the tangent lines and those are in red the tangent
lines are in red is equal to the slope of the line through the end points
which is the slope of the secant line through the endpoints
and so you can see those are equal because all
all those three lines are parallel to each other so they have the same slope
okay so there’s the mean value theorem and now we’re going to work out lots of
examples with it but before we do that i want to do one last thing i want to
make sure i talk to you about how rolle’s theorem and the mean value theorem
are related to each other so i wanted to put them right side by side
let’s read them both of them start off the same
let f be a continuous function on a closed interval
so in that way they’re exactly the same and then
we’re differentiable on the open interval okay they both have
exactly that and roll’s theorem has one more assumption

00:33
and the values that the endpoints are the same
f of a equals f of b whereas the mean value theorem
f a and f b don’t have to be the same but they both say the conclusion
there exists a real number c we’re only talking about real numbers here so
a number c in the open interval a b such that and then they both start off
with the derivative at c is equal to now in the mean value theorem
we don’t have to have f of a equals f of b
but in the case where they are the same then that derivative is zero
so that’s why you can see that rolle’s theorem is a special case of the mean
value theorem the mean value theorem if you prove the
mean value theorem holds then you actually also prove
rolle’s theorem holds because rose theorem is a special case with one more

00:34
assumption all right and so then here’s the pictures for them
so here we have rolle’s theorem and notice that f of a equals f of b they
have the same height and we have a horizontal tangent line
i mean value theorem they don’t have necessarily the same height
so on this graph the f of a is greater than the f of b
but it could be that f of b is greater than f of a so or they could be equal
so it the mean the mean value theorem doesn’t say
so the line can be slanted down like it is in blue
or the line can be slanted up or the line could be horizontal like it is in
rolls theorem but in either case you get a c where the slope of the tangent line
in red is equal to the slope of the line through the end points which is in blue
okay so i hope that you get that rolls theorem is a special case

00:35
of the mean value theorem now let’s go and look at some examples
of the mean value theorem here and practice using it
so we want to find all number c in this interval
where the derivative is equal to the slope of the line through the endpoints
and we’re given a specific function f of x equals 1 plus 1 over x
and a pacific interval here 1 4. so i’m going to take a peek at the graph there
there’s the interval 1 4 and the graph one plus one over x that’s in black
and then we have the slope of the line through the end points
which is the upper line and then we have this uh slope of the tangent line
and so it looks like we’re going to be able to find a c
but you know aside from the graph let’s work on the problem here so
how would we solve this problem using the mean value theorem so first thing is i

00:36
would notice that f is continuous [Music] on one four
and if you wanted to justify that you could say well f of x looks like this
f of x is what x plus 1 over x that’s the same thing as 1 plus 1 over x
in other words this is a rational function rational functions are continuous on
their domain and the domain is all real numbers except
zero zero’s not in one four so this function right here is continuous on 1 4
and it’s continuous from the left and from the right at 1 and 4.

00:37
so we could go and argue all that f is differentiable on open 1 4
and in fact we can go and find that derivative [Music]
derivative of one is zero and then we have x to the minus one
so the derivative is minus one over x squared right minus one over x squared
so when we’re looking at the derivative here we notice the derivative
is existing the domain of this derivative here is all real numbers
except zero so it’s definitely going to be differentiable on open one four
and there’s gonna be the derivative defined on one four [Music]
so this function right here is definitely going to be differentiable on
one fourth uh definitely be uh yeah so the original function is
differentiable one four and there’s the derivative right there all right so

00:38
um those the two things we need to check for
for uh mean value theorem we checked its continuous on the closed interval
we checked it’s differentiable in the open interval
so we know mean value theorem says [Music] abbreviated mvp mvt so
mean value theorem says there exists a c in 1 4 something between 1 and 4
such that the derivative at c which we found to be
minus one over x squared but we’re plugging in c
where the derivative is equal to the slope of the line through the end points
what is the slope of the line through the endpoints it’s f of 4 minus f of 1

00:39
over 4 minus 1. right there’s my endpoints right there so f of 4 minus f of one
that’s rise over run right and so what does this value come out to be
so this right here is just simply negative one fourth [Music]
that right so this will be one plus one-fourth minus one plus one all over three
and so what is this that’s five fourths minus two or eight fourths over three
and so let’s see here sorry that’s a one plus a four so that’s five fourths
minus two two i’ll write this minus eight fourths

00:40
so that’s what minus three fourths over three so minus one fourth [Music]
all right so this right here coming back to all this
this is minus one over c squared is equal to
all of this which we just worked it out it’s minus one fourth so this is minus
one fourth so we look at minus one over c squared is equal to
we calculated all that it’s minus one fourth so c
looks like to be plus or minus two right if we just look right here c is plus or
minus two but if we also look at the interval right here then
we’re bound between one and four so c is two c is two so here we go

00:41
f is continuous on the closed um interval um zero one so let’s go here
how about let’s put a 1 4 here 1 4 and open 1 4 here
and so we’re going to apply the mean value theorem so we’re going to find the
derivative by finding this derivative right here
and plugging in c into the derivative and it’s going to be equal to the mean
value theorem says equals right here so the slope of the line through the
endpoints which we calculate all that up and we get minus 1 4.
so solving this is equal to that right it looks more solvable when you
say minus one fourth so the c here has to be equal to plus or minus two

00:42
c squared equals four so we we find c is two since minus two is not in here
and then we can look right here on the graph right here at two
that’s where we’re getting our um point right there where the slope of the
tangent line is equal to the slope of the line through the endpoints
and there we go all right so next [Music] all right so solve that
minus one fourth there we find the c and you know when you’re solving that
equation right there you have to always remember to go back
and look at your original uh restriction your original domain
all right so now let’s look at this one right here and this is
arc tangent or tangent inverse and so we’re going to find all number c

00:43
where the f private c the slope of the tangent line
is equal to the slope of the line through the end points
so here we go i’m going to take a quick peek at the graph there and so there’s
a tangent inverse on 0 1 and we’re going to notice that it’s continuous on the
closed interval and it’s differentiable on the open interval
so we’re going to apply the mean value theorem always got to check those two
things first before you can apply the mean value theorem
all right so now i’m going to take my derivative so derivative of tangent
inverse is 1 over 1 plus and i’m plugging in c so don’t write x
squared it’s c squared so i we have a video on the uh
derivatives of inverse functions so that’s also in the playlist so if
you’re not sure how to find the derivative of tangent inverse

00:44
all right so the mean value theorem says the derivative at c
is equal to f of b minus f of a over b minus a so what is my a and what is my b
right so my a is zero and my b is one so f of b is tangent inverse of one
and then minus and then f of a so that’s tangent inverse of zero
over b minus a now tangent inverse of zero is zero tangent inverse of one
is pi over four so we’re getting pi over four over one
or just pi over four and so now we have to go
solve that so we have one over one plus c squared equals pi over four
and if we go try to solve that what is the c so let’s just say this is
one one plus c squared equals four over pi just take
the reciprocal of both sides so c squared is four over pi

00:45
and then minus one and so then the c is square root of four over pi minus one
now four and then divided by three point one four four divided by three
point one four right that’s greater than one so this
would be greater than one minus one this will be positive
so you always have to check plus or minus and for this case we’re zero we’re on
zero one so we’re going to choose the positive
one here and this is a real number right here so that’s 4 over pi minus 1
square root that’ll be our c um and then maybe if you want to write it
like instead of that one write it as a pi over pi
i don’t know maybe you write it as 4 minus pi over pi either way that’s our c

00:46
that we know exists by the mean value theorem so we can go and find that c
there we go um the minus one is not in there so we just need to find the c
there’s our c it says find out number c in that interval
and we just did that all right here we go another case of mean value theorem
and let’s look at this so i’m going to look at this function right here f of x
equals x to the third minus x and that’s the polynomial [Music]
um so that’s going to be continuous and differentiable [Music]
all real numbers so [Music] f is continuous on 0 2 and differentiable

00:47
on the open interval 0 2 [Music] so the mean and the mean value theorem [Music]
applies so it says to illustrate the mean value theorem so
first thing i did was check that we can actually use it the mean value theorem
applies so we have a polynomial on a closed
interval so it’s definitely going to be continuous on the closed interval and
differentiable in the open interval all right so let’s see what the mean
value theorem says um so by the mean value theorem there exists a c
in this open interval 0 to 2 such that [Music] the derivative at c is
equal to so we’re looking at 0 2 so it’s going to be

00:48
f of 2 minus f of 0 over 2 minus 0. [Music] so let’s verify that this works
let’s go find the derivative and let’s go find this number here uh we can find
that number here maybe what is what happens when i plug in 2 so
we plug in 2 into our original function up here so that’ll be 8 minus 2 so
that’s 6. and we plug in zero we get zero
so that’s just three so there has to be some c the mean value theorem says
there has to be some c where the derivative is equal to three
the slope of the line through the endpoints is three
there has to be some c where the slope of the tangent line is three
so what is it let’s go find our derivative so here we go derivative
is three x squared minus one and where is that equal to three

00:49
so we have three x squared equals four we have x squared equals four thirds
and so we have x is plus or minus the square root of so two over square
root of three [Music] so we have two values when we go try to
solve this equation here but looking back at our uh interval here
zero to two we’re only going to choose the positive one
so c is two over square root of three or if you want to rationalize
multiply by square root of three over three so c
is you know square root of three you know two over three

00:50
two square roots of three over three [Music] okay so that is that there’s the c
m in this interval zero to two um where the slope of the tangent line
will be equal to the three and so let’s go on now [Music]
next one next uh next part okay so now we’re going to talk about
two applications of the mean value theorem i’m going to work out two
problems for us here the first one is going to be find this limit
and so this will be an application of the mean value theorem to mathematics
itself um in the next example we’ll see applying the mean value theorem to
say a real world problem now the mean value theorem is

00:51
a very important theorem for mathematics it’s used a lot
it’s used to come up with the existence of something
um in this case we’re going to use it to evaluate this limit now this limit can
be evaluated different ways it can be evaluated using a method that we don’t
know about right now uh called la habital’s rule um
so there are other ways to evaluate this limit but i wanted to illustrate how to
use the mean value theorem by finding this limit here for us i
think that that would be fun so let’s see how to do that
so i’m going to use the mean value theorem now to use the mean value
theorem we need a function right they gave us a limit
we take a limit of a function so you might be tempted to take
the whole expression that you’re taking the limit of

00:52
as your function but after doing some scratch work that doesn’t work
so to see what i’m talking about you might be tempted to say this is my
function that i’m going to apply the mean value theorem for
which will be 1 plus x to the n minus 1 over x now the reason why
this is not useful to take this as our function
is because the derivative of this is not going to have any
relevance for this problem and so you know the mean value theorem says f
prime at c is equal to f of b minus f of a over b minus
a so we’re gonna have to look at the if we choose this is our f of x
we’ll have to take the derivative of it and that derivative of this
expression right here doesn’t really have much meaning for this problem right
here so i’m not going to choose that to be my function
instead i’m going to choose now you can apply the mean value theorem to any

00:53
function you want we need to find a useful function that solves this limit
it turns out the function that we want to use is f of x equals 1 plus
x to the n and we want to look on 0x we want to look on 0x so
this is a function it is a polynomial function of degree n
you expand it all out right and we’re looking on it on this interval right here
in fact this is a polynomial so it’s it’s continuous on the whole real line
it’s differentiable on the whole real line we can look at it on any interval
we want i’m going to look on this interval here [Music]
okay so notice that f is continuous and differentiable um

00:54
yeah so that should be written out better so f
is continuous i’ll write it out over here if it’s continuous on 0x
f is differentiable on open 0 to x okay and then we’re going to apply the
mean value theorem so those are the two things that we need to check we need to
we need a function we need a closed interval we check that it’s continuous
we check this differentiable so mean value theorem holds [Music]
and mean value theorem is going to give us the value of this limit
and so we’re going to see that in a minute so what does mean value theorem say
well what’s the derivative of this the derivative is

00:55
n times one plus x to the n minus one and and then times the derivative here
which is just one so there’s the derivative so the derivative at c is equal to
f of the x minus f of the zero over x minus 0. so this will be
the end point right here 0 and x so f of x minus f of 0 and then over x minus 0.
and so if we have f of x that’s our uh one plus x to the n so this right here is
one plus x to the n minus f of zero what is f of zero when we plug in zero
we’re going to get one to the n so we get a one and we get over x [Music]

00:56
so this is why i chose the original function to be 1 plus x to the n
it’s because when we go and found the expression right here that’s what we
want to take the limit of we’re trying to find this limit up here
and now we just need to take the limit of
both sides as x approaches zero we need to take the limit as x approaches zero
but there’s no um x over here it’s f prime at c
so what is f prime at c so if i plug in c in for my x i’m going to get
n times 1 plus c and then n minus 1. so that’s my derivative at c
right there and that is equal to um well if we put a c over here
because if i plug in c over here all right so it’s better to think like this

00:57
um i want to take the limit with respect to x goes to zero here but
there’s no x’s over here so let’s think about this here for a minute
um this interval here is [Music] um [Music] yeah so we have f prime at c
is that expression there where we plugged in c
yes we found all of that right there f prime of c
is equal to and then we have the expression 1 plus x to the n minus 1 over x
so that’s good now this c is on the interval 0 to x so
if we take the limit as x goes to 0 what’s happening to c
right so if we you know have an interval here 0 to x

00:58
and c is in between well x is going to zero so as x goes to 0 c goes to 0
also in other words as x is going to zero c is always between zero and x
so as x gets closer to zero c is also getting closer to zero
so if i take the limit of this side with as x goes to zero
that’ll be the same or that’ll be equal to if we take the limit of this side as
this goes to zero so what we’re getting now is that the limit as c goes to zero
of the derivative at c which is n one plus c to the n minus one
and that will be equal to the limit as x goes to zero of this part right here
so i’m just taking the limit of both sides
here i’m taking x goes to zero and here i’m taking
c goes to zero so c goes to 0 right here and i plug in c
for my x and then over here i have a limit so this will be 1 plus x to the n

00:59
minus 1 over x so this is the limit we’re trying to find
but this is the limit that we know because what is this limit
this is c going to zero here so this is zero so that’s
one to the n minus one this is just n so this limit is just n [Music]
all right so let’s finish writing that up so we found the derivative at c by
finding the derivative and plugging in c and the mean value
theorem says that that is equal to f of b minus f of a over b minus a
which if we plug in the b’s and the a’s into our function
and then we simplify so we have continuity we have differentiable we
know c must exist and if we take the limit as x goes to zero
because c is always in between the two we know that c must also go to zero

01:00
and so the mean value theorem says that c is in that interval always
so we can find the limit the limit of the expression on the far right side
is equal to the limit on the far left side which is equal to n so we found that
limit right there and you might remember back when we were finding limits
we’re doing things like factoring and conjugate
and we’re doing whatever it took to cancel something
because you know when you first look at this limit right here
as x goes to zero and you try to substitute in zero you’re getting zero
over zero and so you know you need to do something and one
thing you don’t want to do is try to multiply all that out but to be honest
you could have done that as well you could have expanded all that out
um the ones would have cancelled you could have cancelled some x’s

01:01
in any case there’s the use of the mean value theorem
so now let’s review a little bit so i’m calling this f of b minus f of a
over b minus a as the average rate of change but if you go back to that episode
um on differentiation we also talked about velocities so this was the average
velocity between time t equals a and t equals b
and the velocity at t equals c at a one particular value in time was called
the instantaneous velocity and so the mean value theorem
if you interpret it in terms of velocities it tells us that if your function is
continuous on the closed bounded interval if your
function is differentiable on the open interval
then there’s somewhere in that interval where the instantaneous rate of change
is equal to the average velocity instantaneous velocity is equal to the

01:02
average velocity at some c not everywhere that some c
all right so here’s an example um i i love this example to be honest
um something that happens to me in real life
perhaps i’ll just say perhaps um okay so two stationary patrol cars
equipped with radar are 1.2 miles apart on the street
for example a street like in arlington texas called cooper street
in any case we have two patrol cars they both have radar and they’re exactly
1.2 miles apart now a van passes the first patrol car
and the speed is 35 miles per hour which may or may not have been the speed
limit but the van is going 35 miles an hour

01:03
one and a half minutes later when the van passes the second patrol car
it’s actually going slower only at 30 miles an hour
so at the one patrol car 35 miles an hour
a little bit down the street it’s going 30 miles per hour
what do you think happened all right so prove that the van must
have exceeded the speed limit of 35 miles per hour at some point
during the one and a half minutes so here’s how we do that so we’re gonna
let t not be the time when the van passes the first patrol car
right so as it passes the first patrol car time started right time is zero
all right so the time it passes the second patrol car
right so it’s one and a half minutes later but i want to put that in terms of
hours because our rates of change are per hour

01:04
35 miles per hour and 30 miles per hour so let’s change those minutes to hours
so now let’s represent s of t to re represent the distance so when i first
started at the first patrol car i mean when the van started
the at the time zero right the distance was zero so s of zero is zero
and then s of that amount of time that’s passed
was 1.2 miles right they’re separated by 1.2 miles so we know those values there
so we can find the average velocity the average velocity is
s of the final moment in time minus s of the original moment in time divided
by the time interval this will be the average velocity
and so if you calculate those up 1.2 divided by the
1.5 divided by 60 and then that gives us the 48 and then i’ll go ahead and put

01:05
the units on here this is the average velocity this is 48
miles per hour right the numerator is 1.2 miles
and then the denominator is in hours so this gives us the average velocity of
48 miles per hour i know what you’re thinking right now
the average velocity was 48 miles per hour so you know if someone is going 35
and then 30 but their average velocity was 48
well the mean value theorem has to apply and why well we’re going to assume the
position function is differentiable and what that means is when you know if
it’s differentiable it’s also continuous right so we’re going to assume it’s
differentiable and then the mean value theorem applies
and then that says that the van must have been traveling
at some point in the interval the the instantaneous rate of change had to
have been equal to the average rate of change

01:06
so somewhere in that somewhere in that drive that 1.2 miles
i had to have gone at least one moment in time
where i had to have gone 48 miles per hour
i mean the van must have been traveling at 48 miles per hour sometime during
this because the instantaneous rate of change
has to be equal to for some value of c okay so you know
question you might ask is hey can you take
the mean value theorem with you into a court of law
uh to uh for your benefit or against you know is that used in the court of
law that’s an interesting question all right next part [Music]
okay so now i want to go over some exercises

01:07
over the mean value theorem and roll’s theorem and
so i want to mention that if you give these a try you know the difference
between an exercise and an example is an example i’m going to work out for you
whereas an exercise is something i want you to try first so
you try these uh problems here and see how they go for you
and if you have a question or something put it in the comments below
and i’ll get back to you and make a video if you if you
if you want that so exercise one two and then we have um
looks like exercise three is missing there so exercise three is not there but
anyways exercise one two four five and six are there
so there’s some very interesting problems one and two just start off by
warming up with the mean value theorem there where you have x to the two thirds

01:08
and you have natural log of x minus one so we work problems just like that um
we worked a problem just like number six or similar to six
and number five you’ll have to translate into a
function and come up with a function and use calculus for it
there’s 7 8 9 10 and 11. so 10 looks just like a problem we solved earlier
and then so seven eight nine you’re going to have to translate
into some theorem and use the theorem like the intermediate value theorem
or rolls theorem or the mean value theorem um
11 is the limit like we solved earlier um yeah and then the last problem
is a thought problem which will be beneficial to you in an upcoming episode
um and so give that a try and see how that goes and uh yeah so try to give those

01:09
exercises a go and see and see what you say leave me some comments below
okay so now i want to say thank you for watching and i hope you’re getting value
out of these videos uh the next video is monotonic functions
and derivative tests and uh so if you like this video please
like and subscribe and if you want to keep up with the
series that i come out with find me on social media in the links below
so thank you so much for watching and i hope to see you next time
if you like this video please press this button
and subscribe to my channel now i want to turn it over to you
math can be difficult because it requires time and energy to become skills
i want you to tell everyone what you do to succeed in your studies
either way let us know what you think in the comments

About The Author
Dave White Background Blue Shirt Squiggles Smile

David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

Let's do some math together, today!

Contact us today.

Account

Affiliates

About Us

Blog

© 2022 DAVE4MATH.com

Privacy Policy | Terms of Service