# 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?

• 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. – Graviton Jul 15 at 10:16
• Also, -3 and 3 are not saddle points as $f'(\pm 3)\neq0$ – Graviton Jul 15 at 10:17
• It's possible that whether they are local extrema would depend on how you have defined "local extrema". – Teepeemm Jul 15 at 19:27

• 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 $$[x_{end}-d, x_{end}+d]$$.

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

• You're correct, but that's completely unrelated to OP's question. – Teepeemm Jul 15 at 19:25

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 $$x=3.$$