Precalculus by Richard Wright

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# 2-09 Nonlinear Inequalities

Summary: In this section, you will:

• Find critical numbers of nonlinear inequalities.
• Solve one-variable nonlinear inequalities algebraically.
• Solve one-variable nonlinear inequalities by graphing.

SDA NAD Content Standards (2018): PC.6.2

###### Solution

This was already rewritten this in example 1 as

$$0 > \frac{-0.386x+5790}{x}$$

and the critical numbers are 0 and 15,000.

Plot those critical numbers on a number line.

Choose a number in each interval between the critical numbers. −100, 5000, and 15100 will work.

Test these in the inequality to see which produce true statements.

$$-100: 0 > \frac{-0.386(-100)+5790}{-100} \rightarrow 0 > -58.286 \text{ True}$$

$$5000: 0 > \frac{-0.386(5000)+5790}{5000} \rightarrow 0 > 0.772 \text{ False}$$

$$15100: 0 > \frac{-0.386(15100)+5790}{15100} \rightarrow 0 > -0.00256 \text{ True}$$

The solutions are the true intervals: (−∞, 0) and (15,000, ∞). Since the hens cannot produce a negative number of eggs, the farm needs to produce more than 15,000 dozen eggs.

Analysis

Both ends of the intervals are parentheses because the inequality is not an equals inequality. If the inequality was ≤ or ≥, there would be a bracket at the end of the interval. However, critical numbers that cause the expression to be undefined always are given a parenthesis in interval notation because x cannot be those numbers.

#### Example 3: Solve a Nonlinear Inequality Algebraically

Solve $$-2 ≤ \frac{2x^2-12}{x}$$.

###### Solution

Solve one side for zero.

$$0 ≤ \frac{2x^2-12}{x} + 2$$

$$0 ≤ \frac{2x^2-12}{x} + \frac{2x}{x}$$

$$0 ≤ \frac{2x^2+2x-12}{x}$$

Find the critical numbers.

$$0 = \frac{2x^2+2x-12}{x}$$

0 = 2x2 + 2x − 12

0 = 2(x2 + x − 6)

0 = 2(x + 3)(x − 2)

x = −3, 2

0 is also a critical number because the inequality is undefined when x = 0.

Graph the critical numbers on a number line and pick test numbers in each interval. −4, −1, 1 and 3 will work for test numbers.

Test the test numbers in the solved inequality.

$$-4: 0 ≤ \frac{2(-4)^2+2(-4)-12}{(-4)} \rightarrow 0 ≤ -3 \text{ False}$$

$$-1: 0 ≤ \frac{2(-1)^2+2(-1)-12}{(-1)} \rightarrow 0 ≤ 12 \text{ True}$$

$$1: 0 ≤ \frac{2(1)^2+2(1)-12}{(1)} \rightarrow 0 ≤ -8 \text{ False}$$

$$3: 0 ≤ \frac{2(3)^2+2(3)-12}{(3)} \rightarrow 0 ≤ 4 \text{ True}$$

The solutions are [−3, 0) and [2, ∞).

Analysis

The critical number 0 comes from when the inequality is undefined, so x ≠ 0. That is why the 0 has a parenthesis by it. The −3 and 2 have brackets by them because x can equal them from the ≤ sign.

##### Try It 2

Solve 0 < 2x2 − 9x − 5.

(−1⁄2, 5)

## Solve Nonlinear Inequalities by Graphing

An alternate way to solve nonlinear inequalities is by graphing. Remember a coordinate plane is simply two number lines set perpendicularly. By solving the inequality so that one side is zero and graphing the expression, the function can be compared with zero. The critical numbers are the zeros and the undefined values. The zeros occur at the x-intercepts; the undefined values often occur at vertical asymptotes. These can both be read from the graph. Intervals on the graph where the y-values are positive are where the inequality is greater than zero. Negative y-values show the intervals where the inequality is less than zero.

###### Solve Nonlinear Inequalities by Graphing
1. Solve the inequality so one side is zero.
2. Graph the resulting expression from the nonzero side of the inequality.
3. If the inequality is > 0, the solutions are intervals with positive y-values.
4. If the inequality is < 0, the solutions are intervals with negative y-values.

#### Example 4: Solve Nonlinear Inequality by Graphing

Solve 8 ≥ x2 − 2x.

###### Solution

Solve the inequality so one side is zero.

8 ≥ x2 − 2x
0 ≥ x2 − 2x − 8

Graph the right-hand expression.

Since the expression is less than zero, the solutions are intervals with negative y-values. The solution is [−2, 4].

#### Example 5: Solve Nonlinear Inequality by Graphing

Solve $$0 < \frac{x-3}{x^2+x-2}$$.

###### Solution

The inequality is already solved for zero, so graph the right-hand expression.

Since the expression is greater than zero, the solutions are intervals with positive y-values. The solutions are (−2, 1) ∪ (3, ∞).

##### Try It 3

Solve $$0 > \frac{x^2}{x-2}$$ by graphing.

(2, ∞)

##### Lesson Summary

###### Nonlinear Inequality

A nonlinear inequality is a mathematical statement containing an inequality such as < or > and the expression does not describe a straight line.

###### Find Critical Numbers
1. Make one side of the inequality equal zero.
2. Find the zeros by replacing the inequality with an equal sign and solve for x.
3. Find any x-values that cause the expression to be undefined, such as dividing by zero or square roots of negative numbers.

###### Solve Nonlinear Inequalities Algebraically
1. Find the critical numbers.
2. Graph the critical numbers on a number line.
3. Choose a number to test in each interval between the critical numbers.
4. Test these numbers in the inequality to see which produce true statements.
5. The intervals that produce true statements are the solutions.

###### Solve Nonlinear Inequalities by Graphing
1. Solve the inequality so one side is zero.
2. Graph the resulting expression from the nonzero side of the inequality.
3. If the inequality is > 0, the solutions are intervals with positive y-values.
4. If the inequality is < 0, the solutions are intervals with negative y-values.

## Practice Exercises

1. When do you use brackets instead of parentheses when writing intervals?
2. Find the critical numbers.

3. f(x) = x2 + 5x − 24
4. g(x) = x3 + 3x2 − 4x
5. $$h(x) = \frac{x^2+4x+4}{x-1}$$
6. Solve inequalities algebraically.

7. x3 − 4x2 − 11x < −30
8. 2 ≥ 6x2x
9. $$0 ≤ \frac{2x}{2x^2+x-1}$$
10. $$1 > \frac{3}{x+1}$$
11. $$0 ≤ \frac{x^2+2x-35}{x-2}$$
12. Solve by graphing.

13. 0 ≥ x2 + 6x + 9
14. 0 < x3 − 7x + 6
15. $$0 ≤ \frac{x}{x^2-3x+2}$$
16. $$0 > \frac{x^2+4x+3}{x-3}$$
17. $$0 < \frac{x+1}{x^3+2x^2-8x}$$
18. Problem Solving

19. It costs a computer software company $13,000 to code a new piece of office software and$2.00 to print and package each disc.
1. Write a function for the average cost of each disk.
2. How many discs do they have to sell to make the average cost less than \$15?
20. Mixed Review

21. (2-08) Graph $$f(x) = \frac{1}{x^2+4}$$.
22. (2-06) Find the zeros of 2x3 + 5x2x − 6.
23. (2-04) Divide (2x3x + 10) ÷ (x2 + 2x − 1).
24. (2-02) Find the vertex of y = 2x2 − 4x + 3.
25. (2-01) Multiply (3 + i)(1 − 3i).

1. Brackets are when the end of the interval is included such as ≤ or ≥. Parentheses are when the end of the interval is not included such as < or >.
2. −8, 3
3. −4, 0, 1
4. −2, 1
5. (−∞, −3) ∪ (2, 5)
6. [−½, ⅔]
7. (−1, 0] ∪ (½, ∞)
8. (−∞, −1) ∪ (2, ∞)
9. [−7, 2) ∪ [5, ∞)
10. −3
11. (−3, 1) ∪ (2, ∞)
12. [0, 1) ∪ (2, ∞)
13. (−∞, −3) ∪ (−1, 3)
14. (−∞, −4) ∪ (−1, 0) ∪ (2, ∞)
15. a. $$C(x) = \frac{2x+13000}{x}$$; b. More than 1000 discs
16. −2, $$-\frac{3}{2}$$, 1
17. $$2x - 4 + \frac{9x+6}{x^2+2x-1}$$
18. (1, 1)
19. 6 − 8i