Solving the Inequality \(1 + \frac{1}{x} - \frac{1}{x^2} \geq 0\)

We are asked to find the solution for the inequality \(1 + \frac{1}{x} - \frac{1}{x^2} \geq 0\), given that \(x \neq 0\).

First, let's combine the terms on the left side of the inequality by finding a common denominator, which is \(x^2\). The inequality becomes:

\(\frac{x^2}{x^2} + \frac{x}{x^2} - \frac{1}{x^2} \geq 0\)

\(\frac{x^2 + x - 1}{x^2} \geq 0\)

Since \(x \neq 0\), the term \(x^2\) is always positive (\(x^2 > 0\)). When we have a fraction \(\frac{A}{B}\) and \(B\) is positive, the sign of the fraction is the same as the sign of the numerator \(A\). Therefore, the inequality \(\frac{x^2 + x - 1}{x^2} \geq 0\) is equivalent to \(x^2 + x - 1 \geq 0\), with the condition \(x \neq 0\) still applying.

Now, we need to solve the quadratic inequality \(x^2 + x - 1 \geq 0\).

To solve a quadratic inequality, we first find the roots of the corresponding quadratic equation, which is \(x^2 + x - 1 = 0\). We can use the quadratic formula to find the roots:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

For the equation \(x^2 + x - 1 = 0\), we have \(a=1\), \(b=1\), and \(c=-1\).

The roots are:

\(x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}\)

\(x = \frac{-1 \pm \sqrt{1 + 4}}{2}\)

\(x = \frac{-1 \pm \sqrt{5}}{2}\)

So, the two roots are \(x_1 = \frac{-1 - \sqrt{5}}{2}\) and \(x_2 = \frac{-1 + \sqrt{5}}{2}\).

The quadratic expression \(x^2 + x - 1\) represents a parabola that opens upwards because the coefficient of \(x^2\) (which is \(a=1\)) is positive. A quadratic \(ax^2 + bx + c\) with \(a > 0\) is greater than or equal to zero (\(\geq 0\)) when \(x\) is less than or equal to the smaller root or greater than or equal to the larger root.

In this case, the smaller root is \(\frac{-1 - \sqrt{5}}{2}\) and the larger root is \(\frac{-1 + \sqrt{5}}{2}\). Therefore, \(x^2 + x - 1 \geq 0\) when:

\(x \leq \frac{-1 - \sqrt{5}}{2}\) or \(x \geq \frac{-1 + \sqrt{5}}{2}\)

We must also consider the original condition that \(x \neq 0\). Let's approximate the roots:

• \(\sqrt{5} \approx 2.236\)
• \(\frac{-1 - \sqrt{5}}{2} \approx \frac{-1 - 2.236}{2} = \frac{-3.236}{2} = -1.618\)
• \(\frac{-1 + \sqrt{5}}{2} \approx \frac{-1 + 2.236}{2} = \frac{1.236}{2} = 0.618\)

The roots are approximately -1.618 and 0.618. Neither of these roots is 0. The solution intervals \( (-\infty, \frac{-1 - \sqrt{5}}{2}] \) and \( [\frac{-1 + \sqrt{5}}{2}, \infty) \) do not include 0 (since 0 is between -1.618 and 0.618). Therefore, the condition \(x \neq 0\) is automatically satisfied by the solution set derived from \(x^2 + x - 1 \geq 0\).

The solution to the inequality \(1 + \frac{1}{x} - \frac{1}{x^2} \geq 0\) is \(x \leq \frac{-1 - \sqrt{5}}{2}\) or \(x \geq \frac{-1 + \sqrt{5}}{2}\).

Comparing this solution with the given options, it matches option 4.

The solution can be summarized as follows:

Revision Table: Key Concepts for Solving Inequalities

Additional Information on Solving Rational Inequalities

Solving rational inequalities generally involves these steps:

1. Move all terms to one side so that the other side is 0.
2. Combine the terms into a single fraction: \(\frac{P(x)}{Q(x)}\).
3. Find the critical points by setting the numerator \(P(x)\) equal to 0 and the denominator \(Q(x)\) equal to 0. The roots of the numerator are where the expression might be zero. The roots of the denominator are where the expression is undefined.
4. Plot the critical points on a number line. These points divide the number line into intervals.
5. Choose a test value within each interval and substitute it into the simplified inequality \(\frac{P(x)}{Q(x)}\) to determine the sign of the expression in that interval.
6. Identify the intervals where the inequality is satisfied based on the desired sign (\(\geq 0, > 0, \leq 0, < 0\)).
7. If the inequality is non-strict (\(\geq\) or \(\leq\)), include the roots of the numerator in the solution if they fall within the intervals, provided they are not also roots of the denominator. Exclude any roots of the denominator from the solution, as the expression is undefined at these points.

In this specific problem, the denominator \(x^2\) was always positive for \(x \neq 0\), which simplified the sign analysis considerably, making the inequality equivalent to just analyzing the numerator.