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- [Instructor] The twice[br]differentiable function G

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and its second derivative[br]G prime prime are graphed.

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And you can see it right over here.

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I'm actually working off of the article

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on Khan Academy called

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Justifying Using Second Derivatives.

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So we see our function G.

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And we see not its first derivative

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but its second derivative here

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in this brown color.

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So then, the article goes on to say,

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or the problem goes on to say,

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four students were asked[br]to give an appropriate

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calculus-based justification for the fact

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

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So let's just feel good[br]that at least intuitively

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it feels right.

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So X equals negative two, remember what

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an inflection point is.

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It's where we're going[br]from concave downwards

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to concave upwards.

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Or, concave upwards to concave downwards.

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Or another way to think about it

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it's a situation where

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our slope goes from[br]decreasing to increasing,

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or from increasing to decreasing.

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And when we look at it[br]over here, it looks like

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our slope is decreasing, it's positive,

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but it's decreasing, it goes to zero.

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Then it keeps decreasing, it becomes

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it's negative now.

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It keeps decreasing until we get to about

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X equals negative two

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and then it seems that it's increasing,

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it's getting less and less[br]and less and less negative.

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It looks like it's a zero right over here

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then it just keeps[br]increasing, it gets more

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and more and more positive.

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

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we go from being concave downwards

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to concave upwards.

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Now a calculus based justification

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is we could look at its,[br]at the second derivative

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and see where the second derivative

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crosses the X-axis.

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Because where the second[br]derivative is negative

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that means our slope is decreasing

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we are concave downwards.

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And where the second[br]derivative is positive

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it means our first derivative

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is increasing, our slope[br]of our original function

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is increasing and we are concave upwards.

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So notice, we do indeed,[br]the second derivative

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does indeed cross the X-axis

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at X equals negative two.

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It's not enough for it to just be zero

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or touch the X-axis,

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

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to have an inflection point there.

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So given that, let's look at the students

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in justifications and see

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what we can, if we can kind of play

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put the teacher hat in our mind

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and say what a teacher would say

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for the different justifications.

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So the first one says,[br]the second derivative of G

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changes signs at X equals negative two.

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

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If the second derivative changes signs

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in this case, it goes[br]from negative to positive,

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that means our first derivative went from

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decreasing to increasing.

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Which is indeed, good for saying this is a

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calculus-based justification.

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So, at least for now I'm gonna put

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kudos you are correct there.

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It crosses the X-axis.

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So this is ambiguous.

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What is crossing the X-axis?

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If a student wrote this I'd say,

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what are they talking about, the function,

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are they talking about[br]the first derivative,

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the second derivative.

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

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This cannot be accepted as[br]a correct justification.

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I will read the other ones.

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The second derivative of G is

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increasing at X equals

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negative two.

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Well no, that doesn't justify

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why you have an inflection point there.

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For example,

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the second derivative is increasing

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at X equals negative 2.5.

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The second derivative is even increasing

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at X equals

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negative one.

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But you don't have an inflection point

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at those places.

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So I would say, this doesn't justify

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why G has an inflection point.

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And then the last student response,

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the graph of G changes concavity

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at X equals negative two.

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That is true

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but that isn't a[br]calculus-based justification.

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We'd want to use our[br]second derivative here.