For \(0 \le x \le 1\), let \(f(x) = \max \{ x^2, 2x(1 - x), (1 - x)^2 \}\). Find the minimum value of f(x) for \(0 \le x \le 1\). Note: For real numbers a, b, c ..., \(max \{a, b, c, \dots\}\) denotes the maximum (or largest) number among a, b, c ....

HexaA May 20, 2021

#1**+1 **

In the above, the functions \(y={x}^{2}\) (blue-green), \(y=2x(1-x)\)(red), and \(y={(1-x)}^{2}\)(purple) are plotted over the interval \(\)[0, 1].

The function f(x) is outlined in black. If I understand correctly, the problem is asking us to find the minimum value of f(x), that is the part of the three graphs outlined in black. The minimum value of f(x) is then \(y={(1-}^{}\frac{1}{3}){}^{2}=\frac{4}{9}\). This was all done on *Winplot.*

Guest May 21, 2021