3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Domain ends are considered critical points because they can have non-zero derivatives yet still be global maximum or minimums. In your case they are only local maximums and minimums. $\endgroup$ – Graviton Jul 15 at 10:16
  • $\begingroup$ Also, -3 and 3 are not saddle points as $f'(\pm 3)\neq0$ $\endgroup$ – Graviton Jul 15 at 10:17
  • $\begingroup$ It's possible that whether they are local extrema would depend on how you have defined "local extrema". $\endgroup$ – Teepeemm Jul 15 at 19:27
5
$\begingroup$
  • 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 $[x_{end}-d, x_{end}+d]$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You're correct, but that's completely unrelated to OP's question. $\endgroup$ – Teepeemm Jul 15 at 19:25
1
$\begingroup$

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 $x=3.$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.