Let's do another problem where
we graph a function based
on the properties of
its derivatives and
second derivative.
And what's especially
interesting about this problem
I think is the fact they
don't give the function.
They just give a bunch of
clues about the function.
So what do they tell us?
Well they tell us that
f of 0 is equal to 3.
They also tell us that at
f of 3 is equal to 0.
They tell us that f
of 6 is equal to 4.
What I like about this problem
is that you can't use a
calculator to solve it because
you don't know the function.
So even if I give you a graphic
calculator you'd be helpless.
Let's write out all the
information then we
can plot it out.
Let me write all the
derivative information here.
They tell us that f prime--
so the derivative of x--
is less than 0 on the
open interval 0 comma 3.
So that means not
including 0 and 3.
And they also tell us that the
derivative of is greater than 0
on the open interval
from 3 comma 6.
So we can already guess that
if we're less than 0 as we
approach 3 and we're greater
than 0 as we go away from 3,
that the derivative
must be 0 at 3.
That's the intuition.
Although they don't tell us
that for sure, but I'm assuming
that this function is
continuous and differentiable
over all the intervals, because
the don't tell us otherwise.
OK.
So that's the information
to tell us about the
first derivative.
And what do they tell us
about the second derivative?
They tell us that the second
derivative is greater than
0 on the interval 0 to 5.
I did the open
interval from 0 to 5.
So that means it's
concave upwards there.
And they also tell us that the
second derivative is less than
0, so it's concave downwards on
the open interval
from five to 6.
You know they don't give
us information what
happens after 6.
Maybe it goes to an
inflection point.
We don't know.
So maybe you know, we'll just
assume that is doesn't change.
So what does all this
information tell us?
Well first let's just graph it.
So f of 0 is 3, so we
have to graph the
point 0,3 3,0 and 6,4.
It seems like all of the action
is in the first quadrant, so
I'll graph it accordingly.
So let me graph it.
I want to focus on the first
quadrant, so that's my x-axis.
That's my y-axis.
I have to go to 6,4.
I want to draw this point
6,4, so 1, 2, 3, 4, 5, 6.
1, 2, 3, 4.
So that's the point 6,4.
And let's see.
3 comma 0.
So that's 1, 2, 3, so that's
this point right here.
That point right
there is 3 comma 0.
I got that from
this information.
And then we have the
point 0 comma 3.
0, 1, 2, 3.
So this is 0,3.
Well, those are definitely
points on the function.
That's what this tells us.
And what do they tell us?
They tell us that the
derivative is less than 0 on
the interval from 0 to 3.
From here to here the
derivative is less than 0.
And that makes sense, because
we have to obviously decrease
to go from this value
to this value.
But what it does tell us is
that the function does not
do something like this.
The function does not do this.
The function does not do that.
And how do we know that?
Well because if the function
moved up and then down, then
you have a positive slope
here, which this says
is not possible.
The derivative is less than 0
from this interval to this
interval, so the slope has to
be negative the entire time.
And then it tells us that the
slope is positive from this
interval-- from x is equal
to 3 to x is equal to 6,
that's what this tells us.
And so you know the line has to
look something like that, or
like that, but can't curve down
then go up, because if it
curved down you'd have a
negative slope at some point.
So OK, fair enough.
Essentially the graph is
monotonic from this point
to this point and this
point to this point.
And if you want to get
technical, the monotonicity
theorem tells us that, well
then this must be a point
either that is a point where
the derivative is 0, or it must
be a point where there is no
derivative, but I'm assuming
that this graph is
differentiable over everything.
And you know you didn't need
a theorem to tell you that.
We know that the slope is
less than 0 as we approach
here in this interval.
We know the slope is greater
than 0 in this interval.
And if we assume that the
derivative is continuous, which
I will make that assumption
although they haven't told us,
but in the absence of that
information, why not.
If the derivative is always
kind of incrementing or
decrementing at a continues
rate, the derivative must be 0
at this point, because it's
less than 0 here and
greater than 0 here.
I'll draw a little horizontal
line to indicate that the
derivative is 0 there.
So a it's going to be a bit
of a minimum point I think.
And then let's move on to the
information on the second
derivatives and see what
they're telling us.
Well they're saying that the
second derivative is greater
than 0 from 0 to 5.
So from 0 to 5.
So this is 5 right here.
What does this tells?
This means that we are
concave upwards from 0 to 5.
That gives us a lot
of information.
So that tells us that not only
are we monotonically decreasing
as we approach 3, but
we're concave upwards.
And and why is this
concave upwards?
Well as you see the
slope here is negative.
It's a little less
negative here.
It's even less negative here
and then as it slowly
approaches 0 it gets less and
less negative, and then
it keeps increasing
until you get to 5.
And then from 5 to
6 what happens?
The second derivative becomes
negative, so we enter a point
that is concave downwards.
So from 5 to 6-- I don't know
what the value of the function
is it 5-- but from 5 to 6 the
function becomes concave
downwards from x equals
5 to x equals 6.
We don't know what f of 5 is,
but f of 5-- they didn't tell
us-- is an inflection point.
That's where we went from
going concave upwards
to concave downwards.
And I think that's all we
can do with this graph.
We don't know what happens.
I mean actually we made a
couple of assumptions anyway,
that the graph is continuous
and differentiable
over the whole thing.
But I think this will give you
an intuition on what all of
this first and second
derivative and all this
stuff does for us.
And just so you know, this
is an inflection point.
they could have told us that f
prime prime of 5 is equal to 0.
We don't know that for sure,
but I'm just assuming.
Anyway hope you found
that vaguely useful.
Actually since I have time,
let me do one more bonus
problem that I was sent.
And it was written right after
that problem so why not do it.
And it says prove that
a quadratic function
has no inflections.
So what does a quadratic
function look like?
f of x is equal to ax
squared plus bx plus c.
And it says prove that this
has no inflection point.
Well, what's an
inflection point?
It's a point where the second
derivative is equal to 0.
So let's figure out what
the second derivative is.
The first derivative of this
quadratic-- but any quadratic
can be written like
this-- is 2ax plus b.
And what's the
second derivative?
Well that's just equal to 2a.
Now an inflection point
is when the second
derivative is equal to 0.
Now my question to you is, can
this function ever equal 0?
Well, the only way that this
function would equal 0 is for
some reason a was 0, but if a
was 0 then it wouldn't be a
quadratic, because then this
term wouldn't exist, it
would be bx plus c and
it would be just a line.
So if we're talking about a
quadratic where this a
coefficient is non 0, which it
has to be because that's the
coefficient of the term x,
there's no way that
this function, this
expression, is non 0.
So there's no way that the
second derivative can ever be
0, and thus we have proved
that a quadratic has no
point of inflection.
see you in the next video.
