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.
