Inequality with absolute value

Task number: 2766

Solve the following inequality over the real domain: \[ \frac{|x+1|}{x-1}\ge x \]

  • Resolution

    The expression is undefined for \(x=1\).

    • \(x\le -1: \frac{-x-1}{x-1}\ge x \longrightarrow -x-1 \le x^2-x \longrightarrow x^2\ge -1 \longrightarrow x\in (-\infty,-1\rangle\)
    • \(-1\le x< 1: \frac{x+1}{x-1}\ge x \longrightarrow x+1 \le x^2-x \longrightarrow x^2-2x-1\ge 0 \longrightarrow \)

      \( \longrightarrow x_{1{,}2}=\frac{2\pm\sqrt{4+4}}{2}=1\pm\sqrt{2} \longrightarrow x\in \left\langle -1, 1-\sqrt{2}\right\rangle\)

    • \(x> 1: \frac{x+1}{x-1}\ge x \longrightarrow x+1 \ge x^2-x \longrightarrow x^2-2x-1\le 0 \longrightarrow x\in \left(1{,}1+\sqrt{2}\right\rangle\)
  • Result

    The solution is \(x\in \left(-\infty,1-\sqrt{2}\right\rangle \cup \left(1{,}1+\sqrt{2}\right\rangle\).

Difficulty level: Easy task (using definitions and simple reasoning)
Routine calculation training
Cs translation
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