Extreme points of a function at domain ends

Question

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 $x_{c_i} = \left\{ -3, -\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}, 3 \right\}$

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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?

Answer 1

• 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 $[x_{end}-d, x_{end}+d]$.

Answer 2

If $D(f)=[-3,3]$, then $f( \pm 3)=0.$

Let $x \in (-3,0)$, then $f(x)<0=f(-3)$, henc $f$ has a local maximum at $x=-3.$

Let $x \in (0,3)$, then $f(x)>0=f(3)$, henc $f$ has a local minimum at $x=3.$

Answer 3

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 $x=-\frac{3}{\sqrt2}.$