Okay, so on a Calculus AB test on telling extrema, concavity, etc. by the 1st and 2nd derivative of a function, we were given a problem as follows.
We were given a graph, labeled $y=p'(x)$. It was piece-wise, but the only important part for this question looks like this:
We were told that there was a vertical tangent at $x=4$. The curve passes through $(4,0)$ and it is clear that when $x<4$, $p'(x)>0$ and when $x>4$, $p'(x)<0$.
Anyways, the question was, where is $p(x)$ concave down. So the way to determine that would be via $p''(x)$, or by just checking where $p'(x)$ is decreasing.
This is what I put:
(something1, 4)U(4, something2]
But according to my teacher, the answer is:
(something1, something2]
He is cool and very open-minded about it and is going to check up on it, and tell me tomorrow. But I'm impatient.
My reasoning is that, because $p''(x)$ approaches $-\infty$ as $x$ approaches $4$, the concavity would be undefined. If we graphed $p(x)$, it would clearly look concave down, but I am convinced that it isn't concave down right at $x=4$.
I figure that all depends on how we define an undefined second derivative. So that's my question. What is the concavity at a point if the second derivative at that point is undefined? My guess would be that the concavity would be undefined.
Best Answer
I would agree with your teacher. The function is concave down on $(-\infty,\mbox{something bigger than 4}]$.
Yes. The second derivative is undefined at $x=4$, but this doesn't negate the possibility of being concave down. The function is concave down if the derivative is decreasing. Agree? Well, looking at your derivative it's decreasing even at $x=4$.
This is related to the fact that a if a function is increasing, we are not guaranteed that the derivative exists (and is positive).
Checking the second derivative is a test for concavity. If the second derivative does not exist, the test does not apply.