Expert: Scotto - 11/24/2008

Question
QUESTION: I understand increasing and decreasing intervals for a function using the first derivative.  Why are the intervals closed and not open when the intervals for concave up and down are open?  It doesn't seem to make sense that you can be increasing and decreasing at the same point.

ANSWER: You're right when you say that it doesn't make sence that a function can be increasing and decreasing at the same point.  The function is increasing at that point or decrasing at the point, not both.  An inflection point is where the function is changing from being concave up to concave down or vice versa.

An inflection point for sin(x) is at 180°.  The function is decreasing on both sides and at this points.  It is in the concave up set for x being in [180°,360°] and in the concave down set for x being in [0°,180°].

The points where the concavity of the function is increasing and decreasing at the same time is called an inflection point.  At inflection points, the second derivative is 0.

If that point is at the end of an interval where the function is concave up, any line drawn from that point to somewhere in the interval is above the curve, so those two points are in the concave up seciton.

If that point is at the end of an interval where the function is concave down, any line drawn from that point to somewhere in the interval is below the curve, so those two points are in the concave down seciton.


---------- FOLLOW-UP ----------

QUESTION: Thanks for responding, but I don't think my questioned was answered.  I totally understand inc/dec and concavity and how to find the intervals.  Both of my calc. books, and all that I have looked up on line give the inc./dec intervals as closed, unless infinity on one side, and the concavity intervals as open- but I have not seen an explanation as to why.  I would say that the concave up of the sine function would be (180,360) [180,360].
Susan

ANSWER: Do you mean (180°,360°) not [180°,360°]?  Well, if 180° is chosen, it is on the edge of the interval.  To be concave up, any line connecting that point to any other point in the iterval must be below the curve.  This is true, so the concavity includes both of the endpoints.  That is the definition of concavity is that if two points on the curve are chosen, the line is on one side and does not cross the curve.  If the lines are below, it is concave down.  If the lines are above, it is concave up.

Go ahead and choose one of the endpoints and connect any line on it to and other point in the section.  You can see that it is below the curve, so it is in the concave up section.

Note also that it is in the concave down section, [0°,180°].  Connect that point on the sin curve to any point in [0°,180°], and you get a line that is below the curve.

That is why it is called an inflection point because it is in both the interval where the curve is concave up and the interval where the curve is concave down.

---------- FOLLOW-UP ----------

QUESTION: Thank you for your responce.  My AP calc. book gives a corollary on inc/dec. fuctions as followes:  Let f be cont. on [a, b] and diff. on (a, b).
1  If f'>0 at each point of (a, b), then f increases on [a, b].
2  If f'<0 at each point of (a, b), then f decresaed on [a, b].
So this means the intervals include the endpoints and a point therefore can be both inc. and dec. at the same time.

The defintion of concavity is as follows:
The graph of a diff. fcn. y = f(x) is
a.  concave up on an OPEN interval I if y' is increasing on I.
b.  concave down on a OPEN interval I if y' is decreasing on I.
So this means that the intervals do not include the endpoints.

The expample of y = sinx, 0<=x<=360, according to my 2 books it has it as increasing on [0, 90] and [270, 360]
and decreasing on [90, 270] and then concave up on (180, 360) and concave down on (0, 180).  So why the difference of intervals?  
Thanks,
Susan

Answer
I went out and found the answer.  It was in
http://www.math.hmc.edu/calculus/tutorials/secondderiv/

It reminds me that to be concave up / down, the line drawn that is tangent at the point must lie below / above the curve in some neighborhood of points around that point.

When you choose and x point that is at the endpoint of the concavity, the line will never be above the curve on (x-δ,x+δ) no matter how small of a δ is chosen.  That is why the endpoints are not included.

To be increasing on [x0,x1], there must be no xa > x0 where
f(xa)≤f(x0).  There must also be no point xa < x1 where f(x1)≥f(x1).
Since this is true even at the endpoints, the interval is taken as [x0,x1].  Even though the slope may be 0 at the endpoint, you can get as close as you want and the inequality is satisfied.

Note that the rules for concavity include a neighborhood around that point whereas the rule for increasing / decreasing only has a one sided test.  That is why concavity is an open interval and increasing / decreasing is a closed interval.