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I did the following problem in class $f(x)=x\sqrt{x+2}$ and I needed to find the local maxima. I said the domain was $[-2,\infty)$ and the left end point $(-2,0)$ on the graph is a local maxima because the graph has a negative derivative there, but my instructor said that it was wrong.

Why is that end point not a local maxima?


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I think the derivative is undefined at $x=-2$. Nonetheless, $f(-2)=0$ and $f(x)<0$ for $x\in(-2,0)$. I'd say there's a local max there. Only possible quibble would be if your definition of local max requires definition in a neighborhood of the point; but, that's rare. – Mark McClure May 9 '13 at 22:19

At the point $(-2, 0)$, $f'(x)$ is not defined. So you reason that the derivative there is negative is incorrect. However, endpoints are usually considered "critical points" which may be candidates for a local extrema. Endpoints are not necessarily local extrema though.

In this case, the endpoint $(-2, 0)$ is a local maximum, since $f(−2)=0$ and $f(x)<0$ for $x\in (−2,0)$. So it is a local maximumum, but not for the reason you give.

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My reasoning that I wrote down was that I graphed it and the graph started going down from there, and there was nothing to the right, but he says that $(-2,0)$ is not a local maximum because it is an endpoint. – user77094 May 9 '13 at 22:27
Then your teacher must be using the definition of local/global "maxima" which excludes endpoints as potential extrema. That is not always the convention. Many teachers do define endpoints as possible extrema, and on that interpretation, you'd be correct. But from now on, for this teacher, I'd suggest omitting endpoints as potential critical points. Does that make sense? (Just know there is no across-the-board convention that everyone follows, regarding endpoints.) – amWhy May 9 '13 at 22:29
Ug - the different definitions can make life so hard! +1 – Amzoti May 10 '13 at 2:49
@Amzoti I completely agree: it's a matter of how such a point is defined. Hopefully, the OP has learned what/how his/her teacher is defining an extremum, BEFORE a final exam! – amWhy May 10 '13 at 2:51
@amWhy: totally agree - looks like a slow night - time to peruse posts and see if I am able to answer anything and then early bed - I woke up at 4 AM (could not sleep). – Amzoti May 10 '13 at 2:53

Too long for a comment...apparently:

A function cannot have a derivative at a point which has some nieghbourhood (or even some one-sided neighbourhood, like in this case!) where the function isn't defined. So $\,f'(-2)\,$ doesn't exist in this case and that perhaps is the reason told you you were wrong.

Now, the function does have a local maxima at $\,(-2,0)\,$ but the reasons are (1) end points are always extrema points, by definition, and (2) the function clearly begins "going down" as it is negative right-close to $\,x=-2\,$ .

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End points need not be extrema - case in point: $x\sin(1/x)$ for $x > 0$ and $0$ for $x=0$. – Mark McClure May 9 '13 at 22:20
Yes, perhaps some "niceness" conditions must be introduced to claim that, though your function isn't defined at $\,x=0\,$ ...but I think you'd define it as zero there, right? – DonAntonio May 9 '13 at 22:23
I explicitly defined it to be zero at zero on edit. The function can be infinitely differentiable, in fact, and still cross the axis infinitely many times. – Mark McClure May 9 '13 at 22:25

It can have a local maxima only if it is continuous and defined on some interval adjacent to it on both the left and the right. Therefore it could be called an absolute maximum (but not in this case) at a point like that, which is at the edge of the function's domain, but not a LOCAL max/min.

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A local maximum of a function $f$ at a point $x = c$ must satisfy: $f'(x) > 0$ for $x < c$ and $f'(x) < 0$ for $x > c$ for some open interval containing $c$.

In your case, there is no open interval containing $-2$ which is why it cannot be a local maximum.

Refer to this:, and this:

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