# Polynomial and Rational Inequalities

We will use a combination of algebraic and graphical methods to solve polynomial and rational inequalities.

### Polynomial Inequalities

Just as a quadratic equation can be written in the form ax2 + bx + c = 0, a quadratic inequality can be written in the form ax2 + bx + c ? 0, where ? is <, >, ≤, or ≥. Here are some examples of quadratic inequalities:
3x2 - 2x - 5 >0, (-1/2)x2 + 4x -7 ≤ 0.
Quadratic inequalities are one type of polynomial inequality. Other examples of polynomial inequalities are
-2x4 + x2 - 3 < 7, (2/3)x + 4 ≥ 0, and 4x3 - 2x2 > 5x + 7.
When the inequality symbol in a polynomial inequality is replaced with an equals sign, a related equation is formed. Polynomial inequalities can be easily solved once the related equation has been solved.

EXAMPLE 1 Solve: x3 - x > 0.
Solution We are asked to find all x-values for which x3 - x > 0. To locate these values, we graph f(x) = x3 -x. Then we note that whenever the function changes sign, its graph passes through an x-intercept. Thus to solve x3 - x > 0, we first solve the related equation x3 - x = 0 to find all zeros of the function:
x3 - x = 0
x(x2 - 1) = 0
x(x + 1)(x - 1) = 0.
The zeros are -1, 0, and 1. Thus the x-intercepts of the graph are (-1, 0), (0,0), and (1, 0), as shown in the figure at left. The zeros divide the x-axis into four intervals: For all x-values within a given interval, the sign of x3 - x must be either positive or negative. To determine which, we choose a test value for x from each interval and find f(x).We can use the TABLE feature set in ASK mode to determine the sign of f(x) in each interval. (See the table at left.) We can also determine the sign of f(x) in each interval by simply looking at the graph of the function.
 Interval Test Value Sign of f(x) (-∞ -1) f(-2) = -6 Negative (-1; 0) f(-0.5) = 0.375 Positive (0, 1) f(0.5) = -0.375 Negative (1, ∞) f(2) = 6 Positive Since we are solving x3 - x > 0, the solution set consists of only two of the four intervals, those in which the sign of f(x) is positive.We see that the solution set is (-1, 0) (1, ∞), or {x| - 1 < x < 0 or x > 1}.

To solve a polynomial inequality:
1. Find an equivalent inequality with 0 on one side.
2. Solve the related polynomial equation.
3. Use the solutions to divide the x-axis into intervals. Then select a test value from each interval and determine the polynomial’s sign on the interval.
4. Determine the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is ≤ or ≥.

EXAMPLE 2 Solve: 3x4 + 10x ≤ 11x3 + 4.
Solution By subtracting 11x3 + 4, we form the equivalent inequality 3x4 - 11x3 + 10x - 4 ≤ 0.
##### Algebraic Solution

To solve the related equation
3x4 - 11x3 + 10x - 4 = 0,
we need to use the theorems of Section 3.4.We solved this equation in Example 5 in Section 3.4. The solutions are
-1, 2 - √2, 2/3, and 2 + √2,
or approximately
-1, 0.586, 0.667, and 3.414.
These numbers divide the x-axis into five intervals:
(-∞, -1), (-1, 2 - √2), (2 - √2, 2/3), (2/3, 2 + √2),and (2 + √2, ∞). We then let f(x) = 3x4 - 11x3 + 10x - 4 and, using test values for f(x), determine the sign of f(x) in each interval: Function values are negative in the intervals (-1, 2 - √2) and (2/3, 2 + √2).Since the inequality sign is ≤, we include the endpoints of the intervals in the solution set. The solution set is
[-1, 2 - √2] [2/3, 2 + √2], or {x|-1 ≤ x ≤ 2 - √2 or 2/3 ≤ x ≤ 2 + √2}.

##### Graphical Solution

We graph y = 3x4 - 11x3 + 10x - 4 using a viewing window that reveals the curvature of the graph. Using the ZERO feature, we see that two of the zeros are -1 and approximately 3.414(2 + √2 ≈ 3.414). However, this window leaves us uncertain about the number of zeros of the function in the interval [0, 1]. The following window shows another view of the zeros in the interval [0, 1]. Those zeros are about 0.586 and 0.667(2 - √2 ≈ 0.586; 2/3 ≈ 0.667). The intervals to be considered are (-∞, -1), (-1, 0.586), (0.586, 0.667), (0.667, 3.414), and (3.414, ∞). We note on the graph where the function is negative. Then including appropriate endpoints, we find that the solution set is approximately
[-1, 0.586] [0.667, 3.414] or {x|-1 ≤ x ≤ 0.586 or 0.667 ≤ x ≤ 3.414}.

### Rational Inequalities

Some inequalities involve rational expressions and functions. These are called rational inequalities. To solve rational inequalities, we need to make some adjustments to the preceding method.

EXAMPLE 3 Solve: $\frac{x-3}{x+4} \ge\frac{x+2}{x-5}$
Solution We first subtract $\frac{x + 2}{x - 5}$ in order to find an equivalent inequality with 0 on one side:
$\frac{x-3}{x+4}-\frac{x+2}{x-5} \ge 0$

##### Algebraic Solution

We look for all values of x for which the related function
$f(x) =\frac{x - 3}{x + 4} - \frac{x + 2}{x - 5}$
is not defined or is 0. These are called critical values.
A look at the denominators shows that is not defined for x = -4 and x = 5. Next, we solve f(x) = 0:
$\frac{x - 3}{x + 4}-\frac{x + 2}{x - 5}=0$
$(x+4)(x-5)[\frac{x-3}{x+4}-\frac{x+2}{x-5}] = (x+4)(x-5)\cdot 0$
(x - 5)(x - 3) - (x + 4)(x + 2) = 0
(x2 - 8x + 15) - (x2 + 6x + 8) = 0
-14x + 7 = 0
x = 1/2.
The critical values are -4, 1/2, and 5. These values divide the x-axis into four intervals:
(-∞, -4),   (-4, 1/2),   (1/2, 5),   and (5, ∞). We then use a test value to determine the sign of f(x) in each interval.  Function values are positive in the intervals (- ∞, -4) and (1/2, 5). Since f(1/2) = 0 and the inequality symbol is ≥, we know that 1/2 must be in the solution set. Note that since neither -4 nor 5 is in the domain of f, they cannot be part of the solution set.
The solution set is (-∞ -4) [1/2, 5).

##### Graphical Solution

We graph
$y=\frac{x-3}{x+4}-\frac{x+2}{x-5}$ in the standard window, which shows its curvature. By using the ZERO feature, we find that 0.5 is a zero.
We then look for values where the function is not defined. By examining the denominators x + 4 and x - 5, we see that is not defined for x = -4 and x = 5
The critical values, where y is either not defined or 0, are -4, 0.5, and 5.
The graph shows where y is positive and where it is negative. Note that -4 and 5 cannot be in the solution set since y is not defined for these values.We do included 0.5, however, since the inequality symbol is ≥ and f(0.5) = 0. This solution set is
(-∞, -4) [0.5, 5).

The following is a method for solving rational inequalities.

To solve a rational inequality:
1. Find an equivalent inequality with 0 on one side.
2. Change the inequality symbol to an equals sign and solve the related equation. 3. Find values of the variable for which the related rational function is not defined. 4. The numbers found in steps (2) and (3) are called critical values. Use the critical values to divide the x-axis into intervals. Then test an x-value from each interval to determine the function’s sign in that interval. 5. Select the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. If the inequality symbol is ≤ or ≥, then the solutions to step (2) should be included in the solution set. The x-values found in step (3) are never included in the solution set.

It works well to use a combination of algebraic and graphical methods to solve polynomial and rational inequalities. The algebraic methods give exact numbers for the critical values, and the graphical methods allow us to see easily what intervals satisfy the inequality.

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