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.