The inequality|x||x−1|<1is equivalent to12<x(x−1)2<x2|x−1|2<|x|21<x2(x−1)21<|x|2|x−1|2|x|2>|x−1|2−x<−12−x>−12

To solve the inequality $|x||x-1| < 1$, we will transform it and analyze the different parts.

1. Break Down the Problem

We need to solve the inequality $|x||x-1| < 1$. We can break it down into the following steps:

1. Understand the expression $|x||x-1|$.
2. Find the critical points by solving $|x||x-1| = 1$.
3. Determine the intervals where the inequality holds.

2. Relevant Concepts

The absolute value function will require consideration of two cases:

• When $x \geq 0$
• When $x < 0$

Additionally, we will work with quadratic expressions derived from the inequality.

3. Analysis and Detail

Case 1: When $x \geq 0$

In this case: $|x| = x \quad \text{and} \quad |x-1| = x - 1 \quad \text{(if } x \geq 1\text{)}$

The inequality becomes: $x(x-1) < 1$ $x^2 - x - 1 < 0$

To find the roots, we can use the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2}$

This gives us two critical points: $r_1 = \frac{1 - \sqrt{5}}{2}, \quad r_2 = \frac{1 + \sqrt{5}}{2}$

Since $r_1$ is negative and irrelevant to this case, we consider $r_2$ which is approximately $1.618$.

Case 2: When $x < 0$

In this case: $|x| = -x \quad \text{and} \quad |x-1| = 1 - x$

This gives us: $-x(1 - x) < 1$ Rearranging gives: $x^2 - x + 1 < 0$

This quadratic has no real roots since its discriminant is negative: $(-1)^2 - 4(1)(1) = 1 - 4 < 0$ Thus there are no valid solutions in this range.

4. Verify and Summarize

Now we find the solution from Case 1. We can test intervals:

1. $(-\infty, 0)$: Not in the solution
2. $(0, r_2)$: Valid (between roots)
3. $(r_2, \infty)$: Not in the solution since the quadratic is positive.

Thus, we conclude:

• The solution is $0 < x < \frac{1 + \sqrt{5}}{2}$.

Final Answer

The solution to the inequality $|x||x-1| < 1$ is: $0 < x < \frac{1 + \sqrt{5}}{2} \quad \text{(approximately } 1.618\text{)}.$