Question

$\frac{x+1}{(x-1)^2}<1$

Answer 1

$(-\infty, 0) \cup (3, \infty)$

Answer 2

To solve the inequality $\frac{x+1}{(x-1)^2}<1$, we follow these steps:

1. Rewrite the inequality so that one side is zero: $\frac{x+1}{(x-1)^2} - 1 < 0$.

2. Combine the terms on the left side into a single fraction: $\frac{x+1 - (x-1)^2}{(x-1)^2} < 0$ $\frac{x+1 - (x^2 - 2x + 1)}{(x-1)^2} < 0$ $\frac{x+1 - x^2 + 2x - 1}{(x-1)^2} < 0$ $\frac{-x^2 + 3x}{(x-1)^2} < 0$ $\frac{x(3-x)}{(x-1)^2} < 0$.

3. Find the critical points by setting the numerator and denominator to zero:

   • Numerator: $x(3-x) = 0 \implies x = 0$ or $x = 3$.
   • Denominator: $(x-1)^2 = 0 \implies x = 1$.

4. Use the critical points to divide the number line into intervals: $(-\infty, 0)$, $(0, 1)$, $(1, 3)$, $(3, \infty)$.

5. Test the sign of the expression $\frac{x(3-x)}{(x-1)^2}$ in each interval. Note that $(x-1)^2 > 0$ for $x \neq 1$, so the sign is determined by $x(3-x)$. We need $x(3-x) < 0$.

6. Analyze the sign of $x(3-x)$. This is a downward-opening parabola with roots at $x=0$ and $x=3$. Thus, $x(3-x) < 0$ when $x < 0$ or $x > 3$.

7. The solution intervals are $(-\infty, 0)$ and $(3, \infty)$. The point $x=1$ must be excluded, but it is not in these intervals.

8. Therefore, the solution set is the union of these intervals: $(-\infty, 0) \cup (3, \infty)$.

Final Answer: $(-\infty, 0) \cup (3, \infty)$