Extreme points of a function at domain ends Consider the following function: $f(x) = x\sqrt{9-x^2}$
$\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad $
$f'(x) = \frac{-2x^2+9}{\sqrt{9-x^2}}$ and $D(f) = [-3,3]$ therefore the critical points of the function are $x_{c_i} = \left\{ -3, -\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}, 3 \right\}$

Apparently the points $\{-\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}\}$ are global minumum and global maximum respectively.
But what about the domain ends $\{-3, 3\}$? Are they considered to be saddle points, local minimums, or local maximums and why?

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*The two domain ends are not considered saddle points since per definition for a saddle point the first derivative has to be zero (necessairy but not sufficient). The derivative $f'(x)$ is not defined at the domain ends as can be seen by your formula (division by zero!)


*The two endpoints can however be described as local extrema (local minimum and maximum). This is the case, since a distance d can be found, such that, the endpoint $x_{end}$ is the minimal/maximal value f can take in the interval $[x_{end}-d, x_{end}+d]$.
