Definition of Critical Point at endpoints On a math test, we were instructed to find critical points of the function $f(x) = x\sqrt{30-x^2}$. I calculated where the derivative was 0, at $\pm \sqrt{15}$, and I knew the domain was restricted to $(-\sqrt{30}, \sqrt{30})$. 
I also was aware that the derivative was undefined at those points, but neither of those points was infinity, in a sense. Like, neither of the "endpoints" was an asymptote, $f(\pm \sqrt{30}) = 0$. 
I got this problem wrong because I included neither $\sqrt{30}$ nor $-\sqrt{30}$ in the list of critical points.
What is the concrete definition of critical point? 
 A: 
What is the concrete definition of critical point?

Before giving the concrete definition, it's better to have a rough idea about critical points.

Critical points are points at which an extremum could possibly occur.

Source: © CalculusQuest™
 endpoint: 
Image sources: Solomon Xie's story, http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png
Therefore, it makes sense to includes three types of points in the domain


*

*points at which the derivative of $f$ vanishes (local min/max, points of inflection)

*endpoints of domain

*points at which the derivative of $f$ is undefined (corner, cusp, points of discontinuity)



On a math test, we were instructed to find critical points of the function $f(x) = x\sqrt{30-x^2}$.

The given domain of $f$ is not clearly stated in the question.  The open interval $(-\sqrt{30},\sqrt{30})$ is just OP's perception.  Taking account of OP's comment and of the 3rd paragraph of the question body, it seems that OP has mistaken the domain of $f$, which should actually be the closed and bounded interval $[-\sqrt{30},\sqrt{30}]$.
Find the critical points type by type.


*

*$f'(x) = \sqrt{30 - x^2} - x \, \dfrac{x}{\sqrt{30 - x^2}} = \dfrac{30 - 2x^2}{\sqrt{30 - x^2}}$, so $f'(x) = 0$ iff $x = \pm\sqrt{15}$

*endpoints of the domain of $f$: $x = \pm\sqrt{30}$

*The denominator of $f'(x)$ vanishes iff $x^2 = 30$, i..e the endpoints of the domain of $f$.


Conclusion: The critical points of $f$ are $x = \pm\sqrt{15}, \pm\sqrt{30}$.
Alternative solution 1
Observe that $f$ is an odd function, since it's a product of an odd function $x \mapsto x$ and an even function $x \mapsto \sqrt{30 - x^2}$.  Therefore, it suffices to find the global maximum of $f$.  Since $f$ vanishes at the endpoints and the midpoint of the domain, the global extrema are actualy local extrema.  The fact that $f$ is odd allows us to concentrates on nonnegative real numbers and apply a.m.-g.m.-inequality
$$\frac{a^2 + b^2}{2} \ge ab$$ with $a = x$ and $b = \sqrt{30 - x^2}$.
\begin{align}
\frac{x^2 + (30 - x^2)}{2} \ge& x \sqrt{30 - x^2} \\
x \sqrt{30 - x^2} \le& 15
\end{align}
Equality holds iff $a = b$.
\begin{align}
\text{i.e. } \quad x &= \sqrt{30 - x^2} \\
x^2 &= 30 - x^2 \\
x &= \pm\sqrt{15}
\end{align}
This elementary solution in algebra-precalculus is useful in contest-math since you don't need to use calculus.
Alternative solution 2
It's even simpler to make use of quadratics.  Note that in the right half of the domain, everything is nonnegative, so squaring $f$ won't affect the answer.  As a result, consider $$(f(x))^2 = x^2 (30 - x^2) = -(x^2 - 15)^2 + 15^2 \le 15^2.$$  This gives the same maximizer $x = \sqrt{15}$ as expected.
