I am tutoring a student in first-semester Calculus. He needs to minimize the function $$f(x)=\frac{\sqrt{4+x^2}}{2}+\frac{\sqrt{1+(3-x)^2}}{4}$$ Taking the derivative and setting it equal to zero, we find (after some cleanup) the equation $$\frac{x}{\sqrt{4+x^2}}=\frac{3-x}{2\sqrt{1+(3-x)^2}}$$ At this point, it seems (to me) that the most likely avenue of solution is to square both sides, cross-multiply, and collect like terms. This leads (after some heavy lifting) to the polynomial equation $$x^4-6x^3+9x^2+8x-12=0$$ Solving quartics is not a lot of fun, but in this case we get lucky: plugging in $x=1$ we find that it is a solution, because $1-6+9+8-12=0$. Phew! And with hindsight, we can see that $\frac{1}{\sqrt{4+1^2}}$ and $\frac{3-1}{2\sqrt{1+(3-1)^2}}$ are both equal to $\frac{2}{\sqrt{5}}$.

But it seems unlikely to me that this is the intended method of solution. First of all, the algebra is considerably thornier than what the student has had to deal with prior to this question. Second, solving a quartic equation seems way out of bounds for a first-semester Calculus class; it happens that in this case the coefficients sum to $0$, but I don't think a typical student would be expected to notice that.

For all of the above reasons, I suspect that there is probably an easier way to solve this problem, but I am at a loss. Is there some trick that I am missing?

  • $\begingroup$ Are graphing calculators or graphers allowed? They would make solving these so much simpler. $\endgroup$ – WaveX Apr 10 '16 at 3:21
  • $\begingroup$ @WaveX Sure, you could use a graphing calculator to find the zeros of the polynomial, or to solve the equation; but if you're going to do that, why not use the same technology to find the minimum of $f(x)$ in the first place? $\endgroup$ – mweiss Apr 10 '16 at 3:23
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    $\begingroup$ Solving "cooked" quartics is generally easy in high school if you use the Rational Root Theorem, which is taught in pre-calculus where I am. $\endgroup$ – Edward Jiang Apr 10 '16 at 3:23
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    $\begingroup$ In terms of solving a quartic, then you could either start guessing factors, or use the quartic formula, the latter of which no one would recommend. In terms of solving optimization problems or just problems in general, then it really depends on the level of the education. Again, from my experiences, the problems given in high school are almost always cooked so the answer turns out nice and simple. $\endgroup$ – Edward Jiang Apr 10 '16 at 3:31
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    $\begingroup$ I am not sure how to justify this (maybe it is a coincidence), but $f(x)$ has the same minimum as $g(x)=\frac{4+x^2}{2}+\frac{1+(3-x)^2}{4}$, that is, $f(x)$ without the roots. If you can justify this, than you can study $g(x)$ instead, which doesn't lead to a quartic. $\endgroup$ – Kuifje Apr 10 '16 at 4:09

Inspired by Kuifje's comment on my question, I think I have a (partial) answer to my own question. Kuifje notes:

I am not sure how to justify this (maybe it is a coincidence), but $f(x)$ has the same minimum as $g(x)=\frac{4+x^2}{2} + \frac{1+(3-x)^2}{4}$, that is, $f(x)$ without the roots. If you can justify this, then you can study $g(x)$ instead, which doesn't lead to a quartic.

Motivated by this observation, let's consider a slightly more general question:

Suppose $f(x)=a\sqrt{h(x)} + b\sqrt{k(x)}$ for some functions $h,k$ which we will assume are continuous, differentiable, non-negative, and whatever else we need as we proceed. Let $g(x)=a\cdot h(x) + b\cdot k(x)$, that is, “$f(x)$ without the roots”. Under what conditions are the critical values of $f$ and $g$ the same?

Towards an answer, let $x_0$ be a value such that $g’(x_0)=0$. Then $a\cdot h'(x_0) + b\cdot k'(x_0)=0$, whence $k'(x_0)=-\frac{a}{b}h’(x_0)$ (assuming that $b\ne 0$).

Now we have $f’(x)=\frac{a}{2} \frac{h’(x)}{\sqrt{h(x)}} +\frac{b}{2} \frac{k’(x)}{\sqrt{k(x)}}$, so

$$f’(x_0) = \frac{a}{2} \frac{h’(x_0)}{\sqrt{h(x_0)}} +\frac{b}{2} \frac{k’(x_0)}{\sqrt{k(x_0)}} = \frac{a}{2} h’(x) \left( \frac{1}{\sqrt{h(x_0)}} - \frac{1}{\sqrt{k(x_0)}} \right)$$

and the latter quantity is $0$ if and only if either $h'(x_0)=0$ or $h(x_0)=k(x_0)$.

Putting this all together, we have the following workflow. Given the task of solving $f'(x)=0$, we:

  1. Solve the (simpler) optimization problem $g'(x)=0$.
  2. For each solution $x_0$ of the simpler problem, check whether $h'(x_0)=0$, and whether $h(x_0)=k(x_0)$.
  3. If either of those two conditions is met, then $x_0$ is also a solution to the original problem $f'(x)=0$.
  4. If you can argue on general grounds that $f(x)$ has at most one critical value, then the problem is solved.

In the specific context of the problem I originally asked, we want to find the critical values of $f(x)=\frac{\sqrt{4+x^2}}{2} + \frac{\sqrt{1+(3-x)^2}}{4}$. Instead, we solve the simpler problem of finding the critical values of $g(x)=\frac{4+x^2}{2} \frac{1+(3-x)^2}{4}$. This leads to the simple, linear equation $$x-\frac{1}{2}(3-x)=0$$ whose only solution is $x=1$. We now check that $\sqrt{4 + 1^2} = \sqrt{1 + (3-1)^2} = \sqrt{5}$. So $x=1$ is also a solution to $f'(x)=0$.

Just to be clear, I don't at all think this is a reasonable technique to expect a first-semester Calculus student to learn -- it is far too specialized to be of general use. But it does solve the problem with much less work than my original solution, and also gives some insight into how one can (in Edward Jiang's words) "cook" examples for which the two functions $f$ and $g$ either do or do not share critical values.

(It also generalizes the common technique in which one needs to minimize a distance, and instead minimizes the square of the distance, on the grounds that any value that minimizes the distance between two points will also minimize the square of the distance.)

I still feel as though there must be another, more elementary way of dealing with this question, and I hope someone else can provide some illumination.


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