Local Maxima in this function 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?
Thanks
 A: 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 $\,x=-2\,$ .
A: 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.
A: 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: http://en.wikipedia.org/wiki/Maxima_and_minima, and this: http://mathworld.wolfram.com/LocalMaximum.html.
A: 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.
