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