Instructions \( \frac{x}{x+1}>3 x \)

To solve the inequality \( \frac{x}{x+1} > 3x \), we will first manipulate the inequality to isolate \( x \). 1. Start by rewriting the inequality: \[ \frac{x}{x+1} - 3x > 0 \] 2. To combine the terms, we need a common denominator. The common denominator is \( x + 1 \): \[ \frac{x - 3x(x + 1)}{x + 1} > 0 \] 3. Simplify the numerator: \[ x - 3x^2 - 3x = -3x^2 - 2x \] So the inequality becomes: \[ \frac{-3x^2 - 2x}{x + 1} > 0 \] 4. Factor the numerator: \[ -x(3x + 2) \] Thus, the inequality is: \[ \frac{-x(3x + 2)}{x + 1} > 0 \] 5. To analyze the sign of the expression, we need to find the critical points by setting the numerator and denominator to zero: - Numerator: \( -x(3x + 2) = 0 \) gives \( x = 0 \) and \( 3x + 2 = 0 \) gives \( x = -\frac{2}{3} \). - Denominator: \( x + 1 = 0 \) gives \( x = -1 \). 6. The critical points are \( x = -\frac{2}{3}, -1, 0 \). We will test the intervals determined by these points: \( (-\infty, -1) \), \( (-1, -\frac{2}{3}) \), \( (-\frac{2}{3}, 0) \), and \( (0, \infty) \). 7. Choose test points from each interval: - For \( x = -2 \) in \( (-\infty, -1) \): \[ \frac{-(-2)(3(-2) + 2)}{-2 + 1} = \frac{-2(-6 + 2)}{-1} = \frac{-2(-4)}{-1} = \frac{8}{-1} < 0 \] - For \( x = -0.5 \) in \( (-1, -\frac{2}{3}) \): \[ \frac{-(-0.5)(3(-0.5) + 2)}{-0.5 + 1} = \frac{0.5(1.5)}{0.5} = 1 > 0 \] - For \( x = -0.1 \) in \( (-\frac{2}{3}, 0) \): \[ \frac{-(-0.1)(3(-0.1) + 2)}{-0.1 + 1} = \frac{0.1(1.7)}{0.9} > 0 \] - For \( x = 1 \) in \( (0, \infty) \): \[ \frac{-(1)(3(1) + 2)}{1 + 1} = \frac{-(1)(5)}{2} < 0 \] 8. The intervals where the expression is positive are \( (-1, -\frac{2}{3}) \) and \( (-\frac{2}{3}, 0) \). 9. We also need to check the endpoints: - At \( x = -1 \), the expression is undefined. - At \( x = -\frac{2}{3} \) and \( x = 0 \), the expression equals zero. 10. Therefore, the solution to the inequality \( \frac{x}{x+1} > 3x \) is: \[ (-1, -\frac{2}{3}) \cup (-\frac{2}{3}, 0) \]