7.6 Solve Rational Inequalities

Solving Rational Inequalities

Rational inequalities ask you to find the set of $x$-values where a rational expression is positive, negative, or compared to some value. Unlike regular equations where you find specific solutions, inequalities produce intervals of solutions. The approach revolves around one key idea: find the critical points that divide the number line, then test each region.

Solving Rational Inequalities

Setting Up the Inequality

A rational inequality contains a polynomial fraction compared to a value, such as $\frac{x+2}{x-1} > 3$. Before you can solve it, you need to get everything on one side so you're comparing a single rational expression to zero.

For example, with $\frac{x+2}{x-1} > 3$:

1. Subtract 3 from both sides: $\frac{x+2}{x-1} - 3 > 0$

2. Combine into a single fraction using a common denominator: $\frac{x+2 - 3(x-1)}{x-1} > 0$

3. Simplify the numerator: $\frac{-2x+5}{x-1} > 0$

Now you have a single rational expression compared to zero, which is the form you need.

Finding Critical Points and Solving

Once the inequality is in the form $\frac{\text{expression}}{{\text{expression}}} > 0$ (or $<, \leq, \geq$), follow these steps:

1. Find the zeros of the numerator. Set the numerator equal to zero and solve. These are the $x$-values where the expression equals zero.
2. Find the zeros of the denominator. Set the denominator equal to zero and solve. These are the $x$-values where the expression is undefined (vertical asymptotes or holes). The expression can never equal zero at these points, so they are always excluded from the solution, even for $\leq$ or $\geq$ inequalities.
3. Plot all critical points on a number line. They divide the line into intervals.
4. Test one value from each interval. Plug it into the simplified rational expression and check whether the result is positive or negative.
5. Select the intervals that satisfy the inequality. Combine them using union notation or inequality notation.

Using the example $\frac{-2x+5}{x-1} > 0$:

• Numerator zero: $-2x + 5 = 0 \Rightarrow x = \frac{5}{2}$
• Denominator zero: $x - 1 = 0 \Rightarrow x = 1$
• Intervals to test: $x < 1$, $1 < x < \frac{5}{2}$, $x > \frac{5}{2}$
• Test $x = 0$: $\frac{5}{-1} = -5$ (negative). Test $x = 2$: $\frac{1}{1} = 1$ (positive). Test $x = 3$: $\frac{-1}{2} = -0.5$ (negative).
• Since we need $> 0$, the solution is $1 < x < \frac{5}{2}$.

Graphing Inequality Solutions

Represent the solution set on a number line:

• Use an open circle (○) for strict inequalities ($<$ or $>$) and for any point where the denominator is zero (always excluded).
• Use a closed circle (●) for inclusive inequalities ($\leq$ or $\geq$), but only at numerator zeros, never at denominator zeros.
• Shade the portions of the number line where the inequality holds.

For the example above ($1 < x < \frac{5}{2}$), you'd place open circles at $x = 1$ and $x = \frac{5}{2}$, then shade the region between them.

If a solution set has multiple intervals, such as $x < -2$ or $x > 1$, shade each interval separately with appropriate circles at the endpoints.

Comparing Rational Functions to Values

Sometimes a problem asks: for what $x$-values is $f(x) > k$? where $f(x)$ is a rational function and $k$ is a constant.

1. Set up the inequality: $f(x) > k$ (or $<, \leq, \geq$).

2. Subtract $k$ from both sides to get $f(x) - k > 0$.

3. Combine into a single fraction and simplify.

4. Solve using the critical-point method described above.

For example, if $f(x) = \frac{x+1}{x-2}$ and you need $f(x) > 5$:

• Rewrite: $\frac{x+1}{x-2} - 5 > 0 \Rightarrow \frac{x+1 - 5(x-2)}{x-2} > 0 \Rightarrow \frac{-4x+11}{x-2} > 0$
• Numerator zero: $x = \frac{11}{4}$. Denominator zero: $x = 2$.
• Test the three intervals, then select where the expression is positive: $2 < x < \frac{11}{4}$.

Properties of Rational Functions

These properties come up repeatedly when working with rational inequalities:

• Domain: All $x$-values where the denominator is not zero. When solving inequalities, domain restrictions tell you which critical points must always get open circles.
• Range: The set of all output values the function can produce. Comparing $f(x)$ to a value $k$ that falls outside the range will yield no solution (or all real numbers in the domain).
• Continuity: A rational function is continuous everywhere in its domain. This is why the sign-test method works: the expression can only change sign at a critical point (numerator zero or denominator zero), so testing one point per interval is enough to determine the sign of the entire interval.