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?

  • $\begingroup$ 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

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.


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.