www.mathcentre.ac.uk/.../8.11%20Maxima%20and%20minima.mp4

In this video, we're going to be
looking at how curves behave and

how to find specific points on

those curves. Let's begin by
looking at a sketch of a curve.

Now we've got a curve. There's
some noticeable things about it.

First of all here.

The curve turns if we
draw tangent to the curve.

It's level parallel to the X
axis. Here I've exaggerated it a

little bit. The curve is flat
and the tangent is again

parallel to the X axis.

Here the curve turns and
again a tangent is parallel

to the X axis.

Now, one of the things that we
do know is that the gradient of

a curve is given by DY by The
X. What's clear is that each of

these points here.

Here and here.

Because the tangent is flat
parallel to the X axis, the

gradient is zero and so D why
by DX is 0.

Now for points like these where
DY by the X is zero, we have

a special name, we call them.

Stationary

points. If
you like the curve, sort of

isn't moving, it isn't changing.
At that point. The why by DX,

which is the rate of change, is

0. Now two of these I discribed
the curve as turning. I said

there the curb was turning and
here again the curve is turning

so that's going from coming up
to going down and this one is

from going down to coming up.
This one isn't this one is very

different so these two.

Are special types of stationary

points. These two.

Are cold turning

points. So at
turning points divided

by DX is

0. Thing to realize though, is
that if the wide by the X is

zero, that doesn't mean we've
got a turning point. We might

have one of these, that's what
makes life just a little bit

tricky, but nevertheless thing
to remember is that turning

points divided by DX is 0. Now
let's have a look at the

nature of this particular
turning point.

In the immediate area of it.

The curve has no bigger value.
We go up, we reach a value and

then we go down again. So this

point. Is called
a local maximum.

Because in the immediate

vicinity. There is no bigger

value. Now usually we forget
about the local and we just call

it a maximum. Similarly, this
point is called a local.

Minimum. Because again, in this
particular region of it, the

curve goes no lower, so this is
the least value in this area are

local minimum. Again, we forget
the word local quite often.

Unjust use minimum. In essence,
this video is going to be about

finding these points.

Maxima and minima finding these
two points Maxima and minima.

And distinguishing between
them because being able to

find them isn't a great deal
of use unless we can tell them

apart. So we need to be able
to distinguish between a

maximum and the minimum. So
let's begin by exploring what

happens around a minimum
point.

Draw a minimum.

Let's take the Tangent.

At the minimum point, we know
that the gradient of this

tangent is 0 because it's
parallel to the X axis. So DY by

DX is 0.

Let's take a point a
little bit before.

The minimum point and take a
tangent at that point. And here

we see that the Tangent.

Is negative, it's got a negative
slope and so D. Why by DX is

negative. I'm using this as a
shorthand for negative, not

actually bothered about the
value of the why by DX just it's

sign. Let's take a point here a
little bit after.

What's happened there? Well, we
can see we've got a line that's

got a positive slope to it,
which means that the why by the

X has got to be positive.

So what's happening as we pass
through this point as X

increases? What's happening to
do, why by DX?

Well, as X increases, the why by
DX goes from negative.

0. To positive.

So divide by the ex is
increasing as X increases, so

let's write that down deep. Why
by DX is increasing.

As X.

Increases. What
does that mean? DY by DX is

increasing well that means that
the rate of change.

Of the why by DX?

Is
positive.

IE D2Y by
DX squared is

greater than 0.

So if D2Y by the X squared is
greater than 0.

And divide by the X is equal
to 0. We know that we have

got a minimum point.

Is that true for all
minimum points?

Little trick of logic here, but
just write down what we've got

if. DY by DX

equals 0. And.

D2Y by DX squared
is greater than

or positive then.

We have

a.
Minimum.

Does it work the other way
around? If we have a minimum?

Do these two both apply? Well,
certainly that one does that if

we have a minimum DY by the X
will be 0, but not necessarily

this one. Let's have a look
at the curve Y equals X to

the 4th. If I sketch this curve.

It looks.

Like that?

Here we've clearly got our
turning point. Our minimum

point. Let's have a look what's

happening. DY by the X equals
well. Why was X to the 4th so

DY by DX is 4X cubed, and we
know that for our stationary

points we put it equal to 0,
which tells us X equals 0 and

X equals 0 tells us that Y
equals 0, so we know that this

is a stationary.

Point

What kind of stationary point is
it? Let's do our second

derivative. D2 why by DX
squared is 12 X squared, but

it equals 0.

When X equals 0.

So this doesn't work in reverse.

If the why by DX is zero and Y2Y
by the X squared is greater than

not, then we know we have a

minimum. But it doesn't work
going the other way. What that

means is that we have to be
especially careful whenever D2,

why? By DX squared comes out to
be 0 when we're doing the test,

so to speak.

We need to look at it in a
slightly different way. In fact,

what we need to do is go back to
basics and I'll show you an

example later where we do go
back to basics and how that

works out. So having had a look
at a minimum point and looked at

it in some detail, let's now
move on and look at a maximum

point and see what happens

there. So here's
our maximum point.

And at the point itself, we know
that the tangent is parallel to

the X axis and so its gradient.

And hence the why by DX is 0.

Let's take a point a little bit
before that there.

We can see that we've got a
line which has a positive

slope to it, and so DY by the
X is positive there.

Now let's take somewhere just
past the point.

And if we have a look at
this week and see we've

got a tangent.

Where the slope is negative and
so divide by the X is negative.

So what's happening as we pass
through this maximum point as X

is increasing? What's happening
today, why by DX?

Well, divide by the axe goes
from being positive to being

zero to being negative. So as
X increases, divided by DX

decreases. As X increases.

Why
by

DX

decreases? Does that mean
it must mean that the rate of

change of the why by DX?

Is.

Negative. As X increases as we
go through this point at this

point. An the derivative
divide by the X is negative.

In other words, D2Y by the X
squared is negative.

So let's just write down again
what we've got.

If. Divide by
DX equals 0.

And. D2 why
by DX squared.

Is less than 0.

Then we have
a maximum point.

Again, just the same argument
applies. Does it work back the

same way? Answer no, it doesn't.
And the problem is what do we do

about D2? Why? By DX squared
being zero? Well, as I said,

will show a way of dealing with
that in a moment. It's time now

to actually look at an example
or two and see if we can make

use of this test.

In order to sort out the
turning points on a curve. So

the first example we're going to
have a look at is Y

equals X cubed, minus three X

+2. The first thing we
need to do is differentiated.

Why by DX equals if we
differentiate this we have three

X squared. We multiply by the
index and subtract 1 prom it

minus the derivative of the
three axis, just three, and this

is equal to 0 for stationary.

Points. We can
factorize this. There's a common

factor of three leaves us with X
squared minus one, and this is

now recognizable as the
difference of two squares, so it

factorizes us three times, X
minus 1X plus one.

And that's equal to 0.

Now what we can see is that.

Either this bracket is 0 or this
bracket is 0, or they're both 0,

so we have X equals 1 or X
equals minus one.

Now this is just given us the X
coordinates of the points that

were interested in what we need
now or the Y coordinates, and so

let's work those out, X equals
1, Y equals and will substitute

it in on the very first line. So
that's one cubed is just 1 - 3

times by one is minus 3 + 2.

Altogether, that gives us 0.

And so our point, our stationary
point is 1 zero. If we

take X equals minus one and
Y is equal to.

And we put minus one in there
well minus one cubed is just

minus 1 - 3 times by minus one
is plus three and plus two at

the end, and altogether 3 + 2 is
5 takeaway, one is 4 and so our

other stationary point is minus

1 four. Now we need to decide.

All these points, Maxima, minima
or that funny one where it was

flat. So in order to sort that
out, what we need to have a look

at is the 2nd derivative.

So I'm going to write the
equation of the curve down

again, and then I'm going to
write down this first

derivative, then the points, and
then the second derivative.

So we've Y equals X
cubed, minus three X +2.

We've got the first derivative
D. Why by DX3 X

squared minus three and the
points that were interested in

with the points one not
and minus 1 four.

So now. Let's take the second
derivative D2Y by DX squared

equals. So we differentiate
this. The derivative of the

first term we multiply by the
index. So two, freezer 6 and

take one of the index that's X
and the derivative of three is

just zero. So let's now look at
the points and. It's the values

of X we're interested in. So X

equals 1. D2Y by
DX squared equals 6.

So this is

positive. What do we
know? We know that if divided by

DX is zero and if D2, why
by DX squared is positive, then

the point we're looking at is a
minimum turning point, so

therefore 10 is a minimum mean

for short. X equals minus
one. This is our second point.

D2Y by the X squared is equal
to minus 6.

And that is negative. The
value of these doesn't

matter. It's their sign
that matters and therefore,

what do we know? We know
that if the Wi-Fi DX is

zero and Y2Y by DX squared
is negative, then we have

got a maximum, so minus 1
four is a Max.

One of the things that this does
help us do is when we do know

the turning points. We can use
this to help us gain a picture

of the graph to help us sketch a
picture. So let's just plot

these two points.

And see if by plotting them we
can build up a picture of the

graph. Now this won't be an

exact picture. The scales will
be different. They'll be a

little bit cramped in some
cases, expanded in others

because what we're trying to
show the important points on the

graph, so we've one and minus
one. Where are two important

values of X, and we know that

one 0. Was one of our turning
points and if we just go back,

we know that one zero was a
minimum turning point, so we

know that at this point the
curve looks like that.

The other point was minus
1 four there.

And we know that at that point
the curve was like that. It was

a maximum. The other thing we
know is that there are no

turning points anywhere else.
These are the only ones, and so

effectively we can join up the
curve and so that we get a rough

picture that it looks something
like that. We can find this

point of course by substituting
X equals 0. Remember, our curve

was Y equals X cubed minus 3X,
plus 2X is 0.

Then this must be 2 here.

This we can find bike waiting
why to zero and solving but

we're not interested in that at
the moment. It's the idea of

using this maximum and minimum
turning points in order to help

us gain a picture of the shape
of the curve.

We're going to have a look at an
example now where the method we

use is one that enables us to go
back to 1st principles, and this

is also the method that we would
use if D2. Why? By DX squared

turned out to be 0.

So it's a very useful method to
have at your fingertips.

So let's begin.

We're going to be looking at
finding the turning points of

this particular curve.

X minus one all squared over

X. Well it's a U over
a V. It's a quotient, so we need

to be able to differentiate as
though it were a quotient. So

let's remember how we do that.
We take the bottom the X.

And we multiply X by the
derivative of U what's on top?

So that's two times X minus one.

Times the derivative of
what's inside a one and

then minus. And we take UX
minus one 4 squared and we

multiply it by the
derivative of E. What's on

the bottom? And that's the
derivative of X. That's

just one, and then it's all
over X squared.

And for our stationary points we
put this equal to 0.

Now.

Let's have a look at this. We
need to tidy it up. It looks a

bit of a mess, so let's have a
look for some common factors.

Well, here we've got an X minus
one and here as well. So let's

take that X minus one out as a
factor. Let's have a look what

we're left with. Well, here
we've got two and X giving us 2X

and one times by one. That's

still 2X. Minus and here we've
got one times by now we just

took out X minus one as a
factor, so it's minus X minus

one and notice I've kept it in
the bracket because it's the

whole of X minus one that I'm

taking away. Close that bracket
all over X squared so this

is still DY by DX.

Now I can simplify what's in the
bracket and have DY by the X is

equal to X minus one times by
now have two X takeaway X that's

just an X, but then I've also to
take away minus one and so that

gives me plus one.

All over X squared and
this is to be equal to 0

for my stationary points.

Now when we have an expression
like that that's equal to 0,

it's only the numerator that
we're interested in. It's only

the numerator that will make
this expression 0.

So got is
X minus 1X

plus one equals

0. And that tells us that
either this bracket is equal

to 0 or that bracket is equal
to 0. In other words, X minus

one equals 0 or X Plus one
equals 0. So X equals 1. Four

X equals minus one.

Now it's the points where after
so let's now calculate the

values of Y, remembering that Y
is equal to X minus one 4

squared all over X.

So let's take X equals 1 and see
what that gives us for why?

We put one in there. We have 1 -
1 all squared over one. Well,

that's just zero because 1 - 1
is 0 square at that still 0 / 1.

The answer is still 0. So one of
our points is one Nord. That's

one of our stationary points.
Now let's do X equals minus one.

Why is equal to minus 1 - 1
all squared all over minus 1 - 1

- 1 is minus 2 and we square
it to that's four but we divide

by minus one, so that is minus
four. So other point is minus 1

- 4. So we've got two
points which are stationary

points and we need to find out
are they turning points and if

they are turning points, are
they Maxima or minima?

So let's write down
again. Why do you? Why

by DX and our points.

So we had why that
was X minus one or

squared over X.

We had DY by the X&YYY DX
was over X squared and let's

just go back so we can see

what. The numerator was it was X
minus one times by X plus one.

X minus one times by
X Plus One and our

two points. Are the
points one note and minus

1 - 4?

Now.

In order to sort out if these
are turning points and if they

are, which is a maximum and
which is a minimum, then we need

to differentiate this again and
that is very very fearsome.

Very fearsome indeed.

Is it really worth
differentiating

differentiating it
again? We might make a

mistake. The algebra
might go wrong.

So rather than differentiating
it again, let's do what we did

before. Let's look at the
gradient a little bit before X

equals 1. At X equals 1 and a
little bit after X equals 1 and

form a picture from doing that
of whether this is a maximum or

minimum point, and this method
that we're going to have a look

at also works.

Wendy to why? By DX squared is
0, so hence it's a good method

and it's well worth adding it to
your Armory of techniques for

looking at stationary points. So
let's set this up and the way

we're going to set it up is too.

Have a little table.

So here's the Y by

The X. And here is
the point where interested in at

X equals 1.

I want to look a little bit
before X equals 1, so I'm going

to say this is X equals 1 minus
Epsilon and all. I mean by

Ipsilon is a little positive bit

of X. North Point One North
Point North one if you like, and

I'm also going to look at the

gradient. A little bit after X
equals 1, so let me call that

one plus Epsilon.

Now I'm not interested in the
value of the why by DX. I'm only

interested in its sign whether
it's positive or negative. I

know that when X equals 1
divided by DX is 0.

What happens if I am a
little bit before X equals 1?

X is a little bit less than one,
so I have something a little bit

less than one, and I'm taking

one away. So the answer to that
is negative. This bracket here

is negative. However, I've got
something that's a little bit

less than one plus one that's

positive. So I have a
negative times by a

positive and I've got X
squared on the bottom here

and X squared is positive,
so I'm negative times a

positive divided by a
positive is a negative.

Let's have a look when X is a
little bit more than one little

bit bigger than one 1.911. Take
away one. Well, a little bit

more than one. We set. Take away
one that's positive. One plus a

little bit more than one that's.

Positive again, and this is
definitely positive in the

denominator, so we've a positive
times by a positive divided by a

positive gives us a positive.

So if we sketch.

The slopes of the tangents. We
have a negative slope aflat

slope under positive slope and
So what we can see is that

this defines the shape of a
minimum and so we can conclude

that one zero is a minimum.

Point notice what we've done is
go back to basics. This is how

we arrived at our test for D2.
Why? By DX squared.

Now let's do it again for this
point, minus 1 - 4.

So we've got DYIDX is equal
to X minus one times by

X Plus one all over X
squared, and the point we're

having a look at is minus
1 - 4. So let's set

up our table again.

We're looking at the sign of DY
by The X.

My looking at it at the point
X equals minus one. As we

pass through that point as X
increases. So we want to know

what's happening alittle bit
before that minus one minus

Epsilon.

And we want to know what's
happening alittle bit after that

minus one plus Epsilon.

Well, we know that when X is
minus one DY by DX is 0. Because

this factor here is 0.

What happens if X is a little
bit less than minus one?

Something like minus 1.1?

Well you got minus 1.1 takeaway.
One that's definitely negative.

Minus 1.1 at on. One that's
minus .1. That's definitely

negative as well.

However, on the bottom here,
we've got a number and we're

squaring it, so it must be
positive. So we've a negative

times a negative, which is a
positive divided by a positive.

So the answer must be positive.

Now let's have a look a little
bit after X equals minus 1 - 1,

plus a little bit.

So the little bit might be not
.1 and minus one plus a little

bit would then be minus N .9. So
we've got minus 9.9 - 1 - 1.9

it's negative. I've got minus
not .9 plus one. Well, that's

not .1, it's positive, so have a
negative times a positive and

this term X squared is always
positive 'cause it's a square.

So we've got a negative times by
a positive and divided by a

positive. So I'll answer. Must

be negative. So let's sketch the
shape. We've got a positive

slope, aflat slope and negative
slope, and this defines the

shape of a maximum, and so
we can say minus 1 -

4 is a maximum.

Finally, let's just see how
this helps us to sketch

the curve. We recall
the equation of the curve

Y equals X minus one
all squared over X, and

we had a maximum point
at minus 1 four.

And we had a minimum
point at one 0.

So some axes, let's mark the .1
and the point minus one goes on

the X axis. Now 10 is here and
we know it's a minimum, so we

know it looks something like
that there minus 1 - 4 down here

somewhere. And we know that
that's a maximum, so it kind of

looks like that.

How can we join these up? Well,
clearly there's something odd

going on around 0 here because
the curve seems to go that way

and that way it seems to go in
opposite directions. What's

happening here? What's going on?
Well, let's just have a little

think about this as X approaches
0 from above.

And this is always going to be
positive, but this is going to

be. Positive as well, and were
divided by something very very

small. Our answer is going to be
very, very big, so we're going

to have something up there when
X is negative but near to 0.

This is still going to be
positive because we're squaring

it, but we're dividing it by
something which is getting very

very small, but negative. So the
answer is going to be very big,

but negative, and so it will be
somewhere down here. So what we

can see. Is that the curve is
going to look something?

Like

That and they will in fact be
this almost what looks like a

sort of hole in the middle of
it. Bang on this value X equals

0. So what have we done?
We have found out how to find

stationary points. Stationary
points occur on a curve. Wendy

why by DX is 0.

If the why by DX is

0? Then some of those points are
what we call turning points,

Maxima or minima.

We can find those.

In one of two ways, if the why
by DX is zero and D2, why by DX

squared is greater than not,
then we know for sure that we

have got a minimum point.

If you divide by the X is zero
and D2 why by DX squared is

negative, then we know for sure
we've got a maximum point. If

however, D2 why by DX squared is
0, then we have no information

whatsoever and we need to look
very closely using the methods

that I've just shown here.

In order to determine what kind
of stationary point we've got,

whether we've got a maximum,
whether we've got a minimum or

whether we've got one of those
odd ones that looked a little

bit flat. A kink in the curve.