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### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 1: AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2011 Calculus AB free response #4c

Finding the points of inflection for a strangely defined function. Created by Sal Khan.

## Want to join the conversation?

- Wouldn't the point (-3, 0) also be a point of inflection, because the rate of change of the slope at that point changes from positive to negative? Just visually looking at the graph leads me to believe this must be the case.(8 votes)
- (-3, 0) is a point of inflection for f(x), but not for g(x). In this problem, the graph shown is the
**derivative**of g(x), so to find inflection of g(x) from**this**graph, we have to see where the slope of this graph goes from positive to negative (or vice versa). I think you're thinking this graph represents g(x), when it is actually g'(x).(10 votes)

- My textbook says that there's more than one definition of inflection point and that the one they use in the book requires that there exists a tangent line and the tangent line intersects the graph. According to my book's definition, would the answer be that g has no inflection points?.(2 votes)
- Yes. This is because this is an odd function, that, like Sal said, is not differentiable at f(0) because the slope jumps. So the book's definition doesn't work for this strange problem.(4 votes)

- I'm a little confused on this. Since f '(x) does not exist at this point, then does this not imply that g ' ' (x) also does not exist? So, therefore, the function has no inflection point?(2 votes)
- Generally... an inflection point is understood to be a point where the concavity/acceleration changes from positive to negative or from negative to positive.

It does not matter whether or not a function's output value exists.

In this case, we just want a point where f'(x) changes between positive and negative.(0 votes)

- In order for a given point to qualify as an
**Inflection Point**of a function**f**, must the second derivative of**f**(with respect to x) be*Continuous*at said point? Or is it simply enough for the function to be**convex**on one side and**concave**on the other (and with a limit that*exists*at that point)?(1 vote)- A point by definition is an inflection point if the concavity changes at that point(from concave up to concave down or vice versa). f does not have to be continuous or differentiable at that point. Nor does the limit have to exist there.(1 vote)

- I've been led to believe that any justification on the AP Calculus AB Exam requires us to use the
**given**information. If that's the case, this would be my justification:

"x = 0 is an extremum point of f(x) and thus an extremum point of f(x) + 2 = g'(x).

x = 0 is, therefore, an inflection point of g(x)."

Would this be okay?(1 vote)

## Video transcript

Part C. Find all values of
x on the interval negative 4 is less than x is less than
3 for which the graph of g has a point of inflection. Give a reason for your answer. So an inflection
point is a point where the sign of the
second derivative changes. So if you take the second
derivative at that point, or as we go close to that point,
or as we cross that point, it goes from
positive to negative or negative to positive. And to think about
that visually, you could think
of some examples. So if you have a curve that
looks something like this, you'll notice that over
here the slope is negative, but it's increasing. It's getting less
negative, less negative. Then it goes to 0. Then it keeps increasing. Slope is increasing, increasing,
all the way to there, and then it starts
getting less positive. So it starts decreasing. So it's increasing. The slope is increasing over
at this point right over here. So even though the
slope is negative, it's getting less
negative over here. So it's increasing. And then the slope
keeps increasing. It keeps on getting
more and more positive up to about this point. And then the slope is
positive, but then it becomes less positive. So the slope begins
decreasing after that. So this right over here
is a point of inflection. The slope has gone from
increasing to decreasing. And if the other thing
happened, if the slope went from decreasing to
increasing, that would also be a
point of inflection. So if this was maybe some
type of a trigonometric curve, then you might see
something like this. And so this also would it
be a point of inflection. But for this, our g of x is kind
of hard to visualize the way they've defined it
right over here. So the best way to
think about it is just figure out where its second
derivative has a sign change. And to think about that, we have
to find its second derivative. So let's write g of x over here. We know g of x is equal to
2x plus the definite integral from 0 to x of f of t dt. We've already taken
its derivative, but we'll do it again. g prime of x is equal to 2
plus-- fundamental theorem of calculus. The derivative of this right
over here is just f of x. And if we have the second
derivative of g-- g prime prime of x-- this is equal to--
derivative of 2 is just 0. And then derivative of
f of x is f prime of x. So asking where this has a
sign change, asking where our second derivative
has a sign change, is equivalent to asking where
does the first derivative of f have a sign change? And asking where the
first derivative of f has a sign change is
equivalent to saying where does the slope of
f have a sign change? You can view this as the slope
or the instantaneous slope of f. So we want to know when the
slope of f has a sign change. So let's think about. Over here, the
slope is positive. It's going up. It's up. It's increasing,
but it's positive. And that's what we care about. So let's write it. I'll do it in green. So the slope is positive
this entire time. It's increasing. It's increasing. It's positive. It's getting less positive now. It's starting to decrease, but
the slope is still positive. The slope is still
positive all the way until we get right over there. You can see it gets
pretty close to zero. And then the slope
gets a negative. And then right over here,
the slope is negative. The slope is negative
right over here. So this is interesting. Because even though f is
actually not differentiable right here-- so f is not
differentiable at that point right over there. And you could see it because the
slope goes pretty close to 0, and then it just
jumps to negative 3. So you have a discontinuity
of the derivative right over there, but we do
have a sign change. We go from having
a positive slope on this part of
the curve to having a negative slope over
this part of the curve. So we experience a sign
change right over here at x is equal to 0, a sign
change in the first derivative of f, which is the same
thing as saying a sign change in the second
derivative of g. And a sign change in the
second derivative of g tells us that, when x is
equal to 0, the graph of g has a point of inflection.