Extreme points of a function at domain ends Consider the following function: $f(x) = x\sqrt{9-x^2}$
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$f'(x) = \frac{-2x^2+9}{\sqrt{9-x^2}}$ 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?

The domain it's $[-3,-3].$
Let $0\leq x\leq 3$.
Thus, by AM-GM $$x\sqrt{9-x^2}=\sqrt{x^2(9-x^2)}\leq\frac{x^2+9-x^2}{2}=\frac{9}{2}.$$
The equality occurs for $$x^2=9-x^2$$ or
$$x=\frac{3}{\sqrt2},$$ which says that $\frac{9}{2}$ is a maximal value of $f$ on $[0,3]$.
The minimal value is $0$ and occurs for $x=0$ or $x=3$.
Let $-3\leq x\leq0$.
Here the maximal value is $0$ and occurs for $x=0$ or $x=-3$.
The minimal value we can get by the similar way:
$$f(x)=-\sqrt{x^2(9-x^2)}\geq-\frac{x^2+9-x^2}{2}=-\frac{9}{2}$$ and it occurs for $x=-\frac{3}{\sqrt2}.$
