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?

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

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.