Solve the compound inequality -1 < x^2 - x - 1 < 1.

Step-by-step solution

1. 1

   Split into two independent inequalities. A chained inequality with a non-linear middle cannot be handled by adding a constant to all three parts and reading off an interval — the middle is not monotonic. It must become two separate conditions joined by "and":

   $x^2 - x - 1 > -1 \quad \text{and} \quad x^2 - x - 1 < 1$

2. 2

   Solve the left inequality. Adding $1$: $x^2 - x > 0$, i.e. $x(x-1) > 0$. A product of two linear factors is positive outside the roots, so

   $x < 0 \quad \text{or} \quad x > 1$

3. 3

   Solve the right inequality. Subtracting $1$: $x^2 - x - 2 < 0$, i.e. $(x-2)(x+1) < 0$. A product is negative strictly between its roots, so

   $-1 < x < 2$

4. 4

   Intersect the two solution sets. Overlaying $\left((-\infty,0) \cup (1,\infty)\right)$ with $(-1,2)$ removes everything outside $(-1,2)$ and the closed block $[0,1]$ in the middle:

   $(-1,\,0) \cup (1,\,2)$

   The gap exists because on $[0,1]$ the parabola dips to a minimum of $-1.25$ at $x = 0.5$, violating the lower bound.

5. 5

   Verify at sample points. At $x = -0.5$: $0.25 + 0.5 - 1 = -0.25$, inside $(-1,1)$. At $x = 0.5$: $-1.25$, outside. At $x = 1.5$: $2.25 - 1.5 - 1 = -0.25$, inside. At $x = 2$: $4 - 2 - 1 = 1$, which fails the strict upper bound, so the endpoint is correctly excluded.