Calculus based justification for an inflection point

- [Instructor] The twice
differentiable function G

and its second derivative
G prime prime are graphed.

And you can see it right over here.

I'm actually working off of the article

on Khan Academy called

Justifying Using Second Derivatives.

So we see our function G.

And we see not its first derivative

but its second derivative here

in this brown color.

So then, the article goes on to say,

or the problem goes on to say,

four students were asked
to give an appropriate

calculus-based justification for the fact

that G has an inflection point
at X equals negative two.

So let's just feel good
that at least intuitively

it feels right.

So X equals negative two, remember what

an inflection point is.

It's where we're going
from concave downwards

to concave upwards.

Or, concave upwards to concave downwards.

Or another way to think about it

it's a situation where

our slope goes from
decreasing to increasing,

or from increasing to decreasing.

And when we look at it
over here, it looks like

our slope is decreasing, it's positive,

but it's decreasing, it goes to zero.

Then it keeps decreasing, it becomes

it's negative now.

It keeps decreasing until we get to about

X equals negative two

and then it seems that it's increasing,

it's getting less and less
and less and less negative.

It looks like it's a zero right over here

then it just keeps
increasing, it gets more

and more and more positive.

So it does, indeed, look
like at X equals negative two

we go from being concave downwards

to concave upwards.

Now a calculus based justification

is we could look at its,
at the second derivative

and see where the second derivative

crosses the X-axis.

Because where the second
derivative is negative

that means our slope is decreasing

we are concave downwards.

And where the second
derivative is positive

it means our first derivative

is increasing, our slope
of our original function

is increasing and we are concave upwards.

So notice, we do indeed,
the second derivative

does indeed cross the X-axis

at X equals negative two.

It's not enough for it to just be zero

or touch the X-axis,

it needs to cross the
X-axis in order for us

to have an inflection point there.

So given that, let's look at the students

in justifications and see

what we can, if we can kind of play

put the teacher hat in our mind

and say what a teacher would say

for the different justifications.

So the first one says,
the second derivative of G

changes signs at X equals negative two.

Well that's exactly what
we were just talking about.

If the second derivative changes signs

in this case, it goes
from negative to positive,

that means our first derivative went from

decreasing to increasing.

Which is indeed, good for saying this is a

calculus-based justification.

So, at least for now I'm gonna put

kudos you are correct there.

It crosses the X-axis.

So this is ambiguous.

What is crossing the X-axis?

If a student wrote this I'd say,

what are they talking about, the function,

are they talking about
the first derivative,

the second derivative.

And so I would say, please
use more precise language.

This cannot be accepted as
a correct justification.

I will read the other ones.

The second derivative of G is

increasing at X equals

negative two.

Well no, that doesn't justify

why you have an inflection point there.

For example,

the second derivative is increasing

at X equals negative 2.5.

The second derivative is even increasing

at X equals

negative one.

But you don't have an inflection point

at those places.

So I would say, this doesn't justify

why G has an inflection point.

And then the last student response,

the graph of G changes concavity

at X equals negative two.

That is true

but that isn't a
calculus-based justification.

We'd want to use our
second derivative here.