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The value of $\displaystyle \underbrace{\min}_{0 \le x \le 1} \max \ [\{x^2, (1 - x)^2, 1 - x^2 - (1 - x)^2\}]$ can be written in the form $\displaystyle\frac{m}{n}$ , where $m$ and $n$ are positive integers with gcd$(m, n) = 1$. Find $m + n$

We can write it as

$\displaystyle f(x)=\max[x^2,(1-x)^2,2x(1-x)]$

Here graph of $f(x)$ as follows

enter image description here

Now solving $\displaystyle x^2=(1-x)^2\Longrightarrow x=\frac{1}{2}$

Also $\displaystyle (1-x)^2=2x(1-x)\Longrightarrow x=1,1/3$

Also $\displaystyle x^2=2x(1-x)\Longrightarrow x=0,2/3$

How I find value of $\displaystyle \min(f(x))$ in $x\in[0,1]$

Please have a look on it

Priti Bisht
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2 Answers2

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enter image description here

The function is symmetrical about $x=\frac12$.

The minimal point is attained at two points, the intersection between $x^2$ and $1-x^2-(1-x)^2$ that is non-zero as well as the intersection between $(1-x)^2$ and $1-x^2-(1-x)^2$ and also that is not $1$. It suffices to study only one of them by symmetry.

$$x^2 = 1-x^2-(1-x)^2, x>\frac12$$

$$2x^2-1 = -(1-2x+x^2), x>\frac12$$

$$2x^2=2x-x^2, x>\frac12$$

Since $x \ne 0$,

$$2x=2-x$$

$$x=\frac23$$

Now, substitute in the equation again, the minimal value is $\left( \frac23\right)^2=\frac49$.

Now you can obtain $m$ and $n$.

Siong Thye Goh
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You need to find the "critical points" inside the interval $(0,1)$. Those are the points where the derivative is equal to $0$, or otherwise does not exist. Your function is, $$ f(x) = \max( x^2, (1-x)^2, 2x(1-x) ) $$

To find the points $p$ where $f'(p) = 0$ you need to differentiate each of the three component functions and set them equal to zero, you will get $p = 0, 1, \frac{1}{2}$.

Next you need to find where the function is not differentiable. This occurs where the three component functions agree. You need to solve the equations $x^2 = (1-x)^2$, and ect. You will find that $p = \frac{1}{3}, \frac{1}{2}, \frac{2}{3}$.

Finally, you need to check the endpoints, but those already been included.

Now when you evaluate you find that at $p = \frac{1}{3}, \frac{2}{3}$ you get the minimum value.

Nicolas Bourbaki
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