The minimum value of the function max {x, x2   } is equal to The minimum value of the function max {x, x2 } is equal to

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$$\begin{gathered} Let f\left(x\right) = max \left\{x, x^2\right\} \\ When x \in \left(0, 1\right), \\ Then, x > x^2. So f\left(x\right) = x \\ When x \in \left(-\infty , 0\right) \cup \left(1, \infty \right), \\ Then, x^2 > x. So, f\left(x\right) = x^2. \\ When x = 0 and x = 1, \\ Then, x = x^2. So, f\left(0\right) = 0 and f\left(1\right) = 1. \\ Thus, f\left(x\right) = \left\{\begin{array}{c}{x}^2, x < 0 \\ x, 0 \le x \le 1\\ x^2, x > 1 \end{array}\right. \\ Now, for x \in \left(-\infty , 0\right) \cup \left(1, \infty \right), f\left(x\right) = x^2 > 0. For x \in 0 \le x \le 1, f\left(x\right) = x \ge 0. Thus, f\left(x\right) is always non-negative. \\ Hence, minimum value of f\left(x\right) is 0 when x = 0. \end{gathered}$$

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