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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?

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  • $\begingroup$ Was the domain restricted to $[-\sqrt{30},\sqrt{30}]$ or $(-\sqrt{30},\sqrt{30})$ ? $\endgroup$ – Lanier Freeman Nov 15 '16 at 8:57
  • $\begingroup$ [ ] @LanierFreeman $\endgroup$ – Saketh Malyala Nov 16 '16 at 1:46
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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™

different types of critical points endpoint: endpoint extremum

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

  1. points at which the derivative of $f$ vanishes (local min/max, points of inflection)
  2. endpoints of domain
  3. 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.

  1. $f'(x) = \sqrt{30 - x^2} - x \, \dfrac{x}{\sqrt{30 - x^2}} = \dfrac{30 - 2x^2}{\sqrt{30 - x^2}}$, so $f'(x) = 0$ iff $x = \pm\sqrt{15}$
  2. endpoints of the domain of $f$: $x = \pm\sqrt{30}$
  3. 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 $$\frac{a^2 + b^2}{2} \ge ab$$ 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 is useful in since you don't need to use .

Alternative solution 2

It's even simpler to make use of . 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.

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