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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 5

Lesson 7: Determining concavity of intervals and finding points of inflection: algebraic- Analyzing concavity (algebraic)
- Inflection points (algebraic)
- Mistakes when finding inflection points: second derivative undefined
- Mistakes when finding inflection points: not checking candidates
- Analyzing the second derivative to find inflection points
- Analyze concavity
- Find inflection points
- Concavity review
- Inflection points review

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# Mistakes when finding inflection points: not checking candidates

Candidates for inflection points are x-values where the second derivative is equal to zero or undefined. But they are only candidates! Once we've found them, we need to test them and see if they really are inflection points.

## Want to join the conversation?

- Just a question: at3:02, since the concavity is positive on both sides, does that mean when x is at 2 that is the minimum?(2 votes)
- yes, you can graph it, by typing:

\left(x-2\right)^{4}

in the left box of this site:

https://www.desmos.com/calculator(2 votes)

- if x = 2 is undefined in the original function, doesn't that discard it as an inflection point as well?(1 vote)
- Correct. Inflection points need to be points in the domain of the original function, similar to how critical points must be in the domain of the original function. If the function itself is undefined at a point, it doesn't make too much sense that it suddenly becomes an inflection point for it.

Plus, for us to even consider inflection points, the curve along with its first and second derivative must be continuous. So, a discontinuity already creates an issue here.(1 vote)

## Video transcript

- [Instructor] Olga was asked to find where f of x is equal to x
minus two to the fourth power has inflection points. This is her solution. So we look at her solution
and then they ask us, is Olga's work correct? If not, what's her mistake? So pause this video and see
if you can figure this out. Alright, let's just follow her work. So here, she's trying to
take the first derivative. So you would apply the chain rule. It would be four times x
minus two to the third power times the derivative of x
minus two which is just one. So this checks out. Then you take the derivative of this. It'd be three times four which would be 12 times x minus two to the second power times the derivative of x
minus two which is just one which is exactly what she has here, 12 times x minus two to the second power. That checks out so step
one's looking good for Olga. Step two, the solution
of the second derivative equaling zero is x equals two. That looks right. The second derivative is 12
times x minus two squared and we wanna make that equal to zero. This is only going to be
true when x is equal to two. So step two is looking good. So step three, Olga says
f has an inflection point at x equals two. So she's basing this just on the fact that the second derivative is
zero when x is equal to two. She's basing this just on the fact that f prime prime of
two is equal to zero. Now, I have a problem with this because the fact that
your second derivative is zero at x equals two, that makes two a nice
candidate to check out but you can't immediately say that we have an inflection point there. Remember, an inflection
point is where we go from being concave upwards
to concave downwards or concave downwards to concave upwards and speaking in the language
of the second derivative, it means that the second
derivative changes signs as we go from below x equals
two to above x equals two but we have to test that because it's not
necessarily always the case. So let's actually test it. Let's think about some intervals. Intervals. So let's think about the interval when we go from negative infinity to two and let's think about the interval where we go from two to positive infinity. If you want, you could
have some test values. You could think about the sign, sign of our second derivative and then based on that, you
could think about concavity, concavity of f. So let's think about what's happening. So you could take a test value. Let's say one is in this interval and let's say three is in this interval and you could say one minus
two squared is going to be, let's see, that's negative
one squared which is one and then you're just going to,
this is just going to be 12 so this is going to be positive
and if you tried three, three minus two squared is one times 12. Well, that's also going to be positive and so you're going to be concave upwards. At least in these test values, it looks like on either side of two that the sign of the second derivative is positive on either side
of the two and you might say, well, maybe I just need
to find closer values but if you inspect the
second derivative here, you can see that this is
never going to be negative. In fact, for any value
other than x equals two and this value right
over here since we're, even if x minus two is
negative, you're squaring it which will make this entire thing positive and then multiplying it
times a positive value. So for any value other than x equals two, the sign of our second
derivative is positive which means that we're
going to be concave upwards and so we actually don't
have an inflection point at x equals two because
we are not switching signs as we go from values
less than x equals two to values greater than x equals two. Our second derivative
is not switching sides. So once again, this is incorrect. We actually don't have an
inflection point at x equals two because our second derivative
does not switch signs as we cross x equals two which means our concavity does not change.