I've been requested to do some problems on critical points and
maxima, minima, and concavity, so I will do some problems.
So these were sent to me by [? Akosh ?].
I don't know where he is but these are interesting problems
and I haven't done anything like these before, so I thought
I would do the exact problem he's given.
So the first problem says, identify the critical point,
and find the maximum and minimum value on the
given interval.
And they say f of x is equal to x squared plus 4x plus 4.
And the interval that they care about is from minus 4, and
including minus 4-- that's why we have the brackets
there-- to 0.
And just so you know, this interval notation, if it was
written like this with parentheses, it would mean all
the numbers between minus 4 and 0, but not including them.
This is a closed interval, so you're including minus
4 and you're including 0.
This is called an open interval.
But anyway you can ignore this, because that's not
what was in the problem.
So the first thing they ask is what are the critical points.
And different people use different terminology, but my
understanding is critical points are all the points
that are interesting.
So it's the points where the derivatives equal to 0, or the
derivative just doesn't exist.
But as we see in this interval, we'll stick to the derivative.
f prime of x is equal to 2x plus 4.
So why does this equal to 0?
Let's see.
2x plus 4 is equal to 0.
2x is equal to minus 4.
I just almost made a mistake there.
x is equal to minus 2.
When x is equal to minus 2, what is f of minus 2?
So we know that f prime of minus 2 is equal 0.
We just solved for that.
So we know the slope is 0 at that point.
But I just want to know what the coordinate point is.
So f of minus 2 is 4 minus 8 plus 4.
So f minus 2 is 0.
Let's evaluate the function at the end points
of this interval.
f of minus 4 is equal to minus 4 squared is 16.
Minus 16, 4 times, so that's 0.
And then plus 4, so f of minus 4 is 4.
And then what's f of 0?
f of 0 is equal to-- well that's 0 0, that's equal to 4.
So now we could graph this.
Let me graph the difference.
So if this is the x-axis and that is the y-axis, and the
interval that we care about is from x is equal to minus 4.
From minus 4 to 0.
So the first thing they said, we'll identify
the critical points.
Well the critical points were when the slope is equal to 0.
And that's when f is equal to minus 2, and that happens at
the point minus 2,0, so it's right here, which is directly
in between those two points incidentally.
Which is a good intuition behind why these two values are
equal, because this is a parabola and it's symmetric, so
if you go 2 away on either side, the function should
have the same value.
But anyway, ignore that if it confused you.
So at minus 2 we have a negative slope.
f of 0 is equal to 4.
And f of minus 4 is equal to 4.
So the graph is going to look something like this.
And I'll do it in another color.
And since we only care about that interval, let's only
graph it over that interval.
It's going to look something like that.
And if you said this was a critical point and they
want to know the maximum and the minimum point.
Well the minimum point is pretty clear.
It's this critical point right here where the derivative
was equal to 0.
And if you wanted to use you know whatever they call
it, the concavity theorem.
But hopefully you have the intuition of why.
You would see that the second derivative at
this point is positive.
And what's the second derivative?
If that's f prime of x, f prime of x for all x is equal to 2.
So really at any point you test you're going to get the second
derivative to be positive.
So especially at this critical point where the first
derivative is 0.
The second derivative being positive tells you that the
slope is constantly increasing.
So the rate of change of the slope is positive.
The slope is really negative there, little less negative,
goes to 0, and then keeps increasing, becomes more
positive, even more positive slope there.
So I think that was out of scope of this exact
problem, but it's good to have that intuition.
And they want to know the maximum points.
Well the value here, x is equal to minus 4 and
the value x equals 0.
They're both maximum points.
They tie for first place.
So those are the maximum points, x equals 0
and x equals minus 4.
You could say 0,4 or minus 4,4.
What's the next problem they give?
Let's see.
Understanding the problem and the terminology is often more
confusing then the problem itself.
So they give h of r is equal to 1/r, and they care about the
interval-- I'm assuming from this interval on r-- from
negative 1 to 3, and we're going to include negative 1 and
3 because we have brackets.
Well what are the critical points?
They're the points where either the derivative is 0, or the
derivative doesn't exist.
Now this brings up an interesting point.
We'll see in a second there's a point here where the derivative
doesn't exist, but the function also isn't defined
at that point.
So that's not a critical point.
So in order to get a critical point-- and this is a
bit of a technicality.
The easy one is derivative is 0.
Derivative equals 0.
The other critical point-- and I was little bit non-rigorous
with it when I said before-- is when there's no derivative.
But f of x is defined.
So as you're going to see in this problem, there is a point
where the derivative is undefined but f of x-- or in
this case, h of r-- is also not defined at that point, so it
won't be a critical point.
Or maybe it is.
I don't know.
Depending on how your teacher defines it.
But the way I learned it is a critical point-- if the
derivative isn't defined, but the function is.
But anyway.
What's the derivative here?
h prime of r.
Well that's r to the negative 1, right, so it becomes
negative r of the negative 2, or that's minus 1
over r squared.
That's the derivative.
Can this ever equal 0?
Well no, it's never going to be equal 0.
But where is it undefined?
Well it's undefined when r is equal to 0.
And r equal to 0 is in that interval.
So h prime of 0 is undefined.
But h of 0 is also undefined, right?
1/0 is undefined.
h of 0 also undefined.
So I'm not going to really consider that to be a
critical point frankly.
It's just a point at which the derivative and the
function is undefined.
And so what will the graph look like?
Well let's see if we can draw the interval.
And we might just want to draw some points.
That's what I always used to do if I never got
stuck on a math problem.
I would just graph some points.
You can ever go wrong with that.
Let's see, 1/r.
So we want to go from negative 1 to 3, so we're going to
have some negative values.
So we want to go from r-- this is the r-axis, right, because
this is the function of r, this is the h of r-axis.
So we're going to go from r is equal to negative
1 to r is equal to 3.
And we could plot some points.
What is h of negative 1?
It's 1 over negative 1, so it's this point right here.
And then what happens is we get to smaller and
smaller negative numbers.
Well think about it.
If you have negative 1/2 here.
When r is negative 1/2, it goes to negative 2, and it's
actually going to asymptote.
We only care about this interval, so it's going to
look something like this.
It's going to asymptote to negative infinity, right, if I
kept drawing this all the way down that would just asymptote.
And then what is h of 3.
Well h of 3 is just going to be 1/3.
It's going to be relatively low number.
And then what's going to happen as we get closer
and closer to 0?
Well if you put a really small positive value here, you get
larger and larger numbers.
So this actually asymptotes to positive infinity, so it'll
look something like this.
I always have trouble drawing these hyperbolas.
It's going to look something like that, it's going to
go to positive infinity.
So that goes to positive infinity, and this will on this
side go to minus infinity.
And frankly just looking at this graph you see why the
graph is not defined at 0.
From the right hand side the limit approaches
positive infinity.
From the left hand side it approaches negative infinity,
and 1/0 is just undefined.
Similarly that's why you couldn't get a derivative there
either because the definition of the derivative means you're
taking a limit and the limit has to be valid from
both directions.
And as we see the limit is not valid from
both directions here.
So what are the maximum and minimum points here?
Well if we look at this value, we might say oh,
is that a minimum point?
Well no, because you have values much lower.
Actually you have values that go to minus infinity
in this interval.
And then you say, well is this a maximum point?
Well no, you have values that go to positive infinity
in this interval.
So it actually turns out that there is no-- I mean this isn't
very proper-- but to some degree at 0 you're at both
positive and negative infinity depending on what direction
you're coming from.
But this really has no maximum or minimum points, because
the graph is undefined at 0.
And then you know if I said, oh well what if I get really close
to 0 and I called that the maximum point.
Then you say, no, no, no, no, but there's a point even closer
that has even a higher value, that's even closer to infinity.
So there's actually no exact point that is
the maximum value.
So that's kind of interesting.
Anyway, I'm already at 11 minutes and people at YouTube
apparently let me go well past the limit now, but people told
me that they like the 10 minute nuggets of knowledge, so I will
stop now and I'll continue these problems in
a future video.
See you soon.