Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to minimize the function $$f(x) = \max \left\{ \frac{a}{x^2}, \frac{b}{1 - x} \right\},$$ where $a, b > 0$ are constants, and $0 < x < 1$.

Is there a way to find an $x$ which will make $f(x)$ as small as possible?

When I plot $\frac{1}{x^2}$ and $\frac{1}{1 - x}$ together, we can see that $x \approx 0.62$ minimizes the maximum of these two functions. Will the answer always be that the minimizing $x$ occurs where $\frac{a}{x^2} = \frac{b}{1 - x}$?

plotting both functions

share|cite|improve this question
Should we also assume $a>0$ $b>0$? – leonbloy Apr 17 '12 at 15:39
To your last question: No the minimum will not necessarily be where the two functions equal. Take for example two decreasing functions. Or take two functions that never intersect. – Thomas Apr 17 '12 at 15:40
@leonbloy Yes, we should assume $a, b > 0$. Sorry for not clarifying; I'll edit the question. – jamaicanworm Apr 17 '12 at 15:40
@Thomas Yes, I know this will not always be the case for arbitrary functions. But for this function $f(x) = \max\{\frac{a}{x^2}, \frac{b}{1-x}\}$, is it true? – jamaicanworm Apr 17 '12 at 15:41
@jamaicanworm Ah... sorry. – Thomas Apr 17 '12 at 15:42
up vote 11 down vote accepted

For simplicity, let's write $g(x) = \frac{a}{x^2}$ and $h(x)=\frac{b}{1-x}$. Then $f=\max\lbrace g,h\rbrace$. Note that $g$ is strictly decreasing and $h$ is strictly increasing. The function $g-f$ is a continuous function such that $\lim\limits_{x\downarrow 0}=+\infty$ and $\lim\limits_{x\uparrow1}=-\infty$. Since $g-f$ is continuous, it therefore must have a zero somewhere on $(0,1)$. (Added: this zero is unique, since $g-f$ is strictly decreasing, as it is the sum of stricly decreasing functions $g$ and $-f$.)

So the two graphs will necessarily intersect in a unique point $x_0\in(0,1)$. Now, $f(x)=g(x)$ for $x\in(0,x_0]$ and $f(x)=h(x)$ for $x\in[x_0,1)$, so $f$ is strictly decreasing on $(0,x_0]$ and strictly increasing on $[x_0,1)$. This implies the only possible point where it might achieve a minimum is $x_0$ and also that it indeed achieves a minimum there. To calculate the point $x_0$, just solve the quadratic equation you get.

share|cite|improve this answer

In general, if you have two functions $f_1(x),f_2(x)$ continuous in $(0,1)$ respectively decreasing/increasing, AND if there exists some $x_0 \in (0,1)$ such that $f_1(x_0)=f_2(x_0)$ (it can exist at most one), then it's obviously true that the minimum of $\max(f_1(x),f_2(x))$ occurs at $x_0$. And this is the case here.

share|cite|improve this answer

If a and b are both positive, consider the derivative of $\frac{a}{x^2}$ and $\frac{b}{1-x}$ being $\frac{-2a}{x^3}$ and $\frac{b}{(1-x)^2}$, respectively. In the range 0 < x < 1, $\frac{-2a}{x^3} < 0$ and $\frac{b}{(1-x)^2} > 0$ throughout. So minimum can occur only at 0, 1, or at the point where these meet. Since $a/x^2$ will tend to infinity as x tends to 0 and $\frac{b}{1-x}$ will tend to infinity as x tends to 1, the minimum will come only at the point where these meet.

If a and b are both negative, a similar argument will work because the max function will tend to 0 at both ends but will be negative elsewhere.

If only one is negative, then the function will basically become the function with the positive coefficient, so the minimum will occur at x tending to 0 (for a positive) or 1 (for b positive).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.