Let , where f_1(x) = cases x^2, |x| , & |x| 1, x^2, |x| , & |x| > 1, cases and f_2(x)…

MATHEMATICS · Single correct · AoPS · Mathematics · Special Functions · Special Functions

Problem

Let f(x)=f1(x)−2f2(x)f(x) = f_1(x) - 2f_2(x), where f1(x)={min⁡{x2,∣x∣},∣x∣≤1,max⁡{x2,∣x∣},∣x∣>1,f_1(x) = \begin{cases} \min\{x^2, |x|\}, & |x| \le 1,\\ \max\{x^2, |x|\}, & |x| > 1, \end{cases} and f2(x)={min⁡{x2,∣x∣},∣x∣>1,max⁡{x2,∣x∣},∣x∣≤1.f_2(x) = \begin{cases} \min\{x^2, |x|\}, & |x| > 1,\\ \max\{x^2, |x|\}, & |x| \le 1. \end{cases} and let g(x)={min⁡{f(t):−3≤t≤x},−3≤x<0,max⁡{f(t):0≤t<x},0≤x≤3.g(x) = \begin{cases} \min\{f(t) : -3 \le t \le x\}, & -3 \le x < 0,\\ \max\{f(t) : 0 \le t < x\}, & 0 \le x \le 3. \end{cases} For x∈(−1,0)x \in (-1, 0), f(x)+g(x)f(x) + g(x) is (1) x2−2x+1x^2 - 2x + 1 (2) x2+2x−1x^2 + 2x - 1 (3) x2+2x+1x^2 + 2x + 1 (4) x2−2x−1x^2 - 2x - 1

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