In mathematics a polynomial inequality can be written as follows:

P(*x*) ≥ 0, (≤ )

where P(*x*) is a reducible polynomial.

**Example**. P(*x*)= *x*^{2} – *x* – 6 ≥ 0

We will solve polynomial inequalities using the following steps:

Rewrite the inequality so that there is a zero on the right side (for example P (*x*) > 0).

Find all polynomial factors of the polynomial function on the right side: P(*x*) = P_{1}(*x*) · P_{2}(*x*) ·… · P_{n}(*x*) > 0

To find the critical values of polynomial function on the right side, set each polynomial factor to zero and solve for *x*.

Determine the sign of each polynomial factor in the intervals delimited by its roots.

The final solution includes intervals, given by the rule sign, in which the polynomial function P(*x*) has the sign satisfying the assigned inequality.

**Exercise 6.** Solve the following inequality:

*x*^{2} – 3 > 2*x*

Solution. Rewrite the inequality with one side equal to zero

*x*^{2} – 2*x* –3 > 0

Find the zeros of the polynomial *x*^{2} – 2*x* – 3 and factor the left side of the inequality:

(*x* + 1)(*x* – 3) > 0

The two real zeros -1 and 3 of the left side of the inequality, divide the real number line into 3 intervals:

(-∞, -1), (-1, 3), (3,+ ∞)

Check the sign of the product (*x* + 1)(*x* – 3) on these intervals:

*x* + 1 > 0 if *x* > -1

*x* – 3 > 0 if *x* > 3

then

*Sign of x* + 1 > 0

*Sign of* *x* – 3 > 0

*Sign of (x *+ 1)(*x* – 3) > 0

In fact

*x* + 1 > 0 ⇒ *x* > -1

*x* – 3 > 0 ⇒ *x* > 3

And then (*x* + 1)(*x* – 3) > 0 if

Finally, the solution in interval notation is

(-∞, -1) ∪ (3,+ ∞)

The graph of solution is sketched below.

**Exercise 7**. Solve the following inequality: − *x*^{2} + 4 < 0

*Solution*. Multiply both sides by -1 and REVERSE (-1 is negative) the inequality sign

*x*^{2} – 4 > 0

Find the zeros of the polynomial *x*^{2} – 4 and factor the left side of the inequality:

(*x* – 2)(*x* +2) > 0

The two real zeros -2 and +2 of the left side of the inequality, divide the real number line into 3 intervals:

(-∞, -2), (-2, 2), (2,+ ∞)

Check the sign of the product (*x* – 2)(*x* + 2) on these intervals:

*x* + 2 > 0 if *x* > –2

*x* – 2 > 0 if *x* > 2

then

*Sign of x* + 2 > 0

*Sign of x* – 2 > 0

*Sign of (x* + 2)(*x* – 2) > 0

In fact

*x* – 2 > 0 ⇒ *x* > 2

*x* + 2 > 0 ⇒ *x* > –2

And then (*x* – 2)(*x* + 3) > 0 if

Finally, the solution in interval notation is

(-∞, -2) ∪ (2,+ ∞)

The graph of solution is sketched below.

**Exercise 8.** Solve the following inequality: *x* (*x*^{2} + 4) > 0

*Solution*. Let us note that the polynomial factor (*x*^{2} + 2) is always positive and the sign of the assigned polynomial depends only on the zero 0 of polynomial factor *x*. The real zero 0 divides the real number line into two intervals:

(– ∞, 0), (0, + ∞ )

Check the sign of *x* on these two intervals.

Finally, the solution in interval notation is

(0,+ ∞)

The graph of solution is sketched below.

**Exercise 9.** Solve the following inequality:

2*x*^{2} + 4*x* ≥ *x*^{2} – *x* – 6.

*Solution*. Verify that the solution in interval notation is

(-∞, – 3] ∪ [- 2,+ ∞)

**Exercise 10.** Solve the following inequality:

*x*^{2} + 2*x* – 8 < 0.

*Solution*. Verify that the solution in interval notation is

[- 4, 2 ]

A rational inequality can be written as follows:

P(*x*)/Q(*x*) > 0 (or P(*x*)/Q(*x*) > 0)

where P and Q are polynomials.

**Example**. P(*x*)/Q(*x*) = (*x* – 2) / (*x* + 3)

The process for solving rational inequalities is nearly identical to the process for solving polynomial inequalities with a few minor differences. The sign of the rational expression P/Q depends on the signs of P and Q. In turn the signs of P and Q depend on the zeros of P and Q respectively if there are any. Hence the sign of P/Q depends on the zeros of both P and Q and changes (if it does!) only at these zeros. So to solve a rational inequality of the form P/Q > 0 (or P/Q < 0), we first find the zeros of P and Q then make a table of sign of P/Q.

**Exercise 11**. Solve the following rational inequality:

Solution. The critical numbers for a rational inequality are all the zeros of the numerator and the denominator. Since the numerator and denominator are already factored in this example, we see that the critical numbers are: -3, 5, and 1.

The three critical numbers divide the number line into four intervals:

(-∞, -3), (-3, 1), (1,+ ∞) 1

Check the sign of the rational function on these intervals:

*x* + 3 > 0 if *x* > – 3

*x* – 5 > 0 if *x* > 5

*x* – 1 > 0 if *x* > 1

then

*Sign of x* + 3 > 0

*Sign of x* – 5 > 0

*Sign of x* – 1 > 0

*Sign of*

And then the rational function

is positive if: – 3 < *x* < 1 and *x* > 5.

Finally, the solution in interval notation is (- 3, 1) ∪ (5,+ ∞)

The graph of solution is sketched below.

Inequalities involving absolute values can be rewritten as combinations of inequalities.

Let *a* be a positive number.

| *x* | < *a* if and only if – *a* < *x* <* a*.

| *x* | > *a* if and only if *x* < – *a* or *x* > *a*.

**Exercise 12.** Solve the following Inequality involving absolute value:

| *x* + 2 | < 3

*Solution*. The above inequality is solved by writing a double inequality equivalent to the given inequality but without absolute value:

– 3 < *x* + 2 < 3

Solve the double inequality to obtain: – 5 < *x* < 1

The above solution set is written in interval form as follows: (– 5, 1)

and the graph of solution is sketched below.

Exercise 13. Solve the following Inequality involving absolute value:

| – 2*x* – 4 | ≥ 9

*Solution*. Solving the above inequality is equivalent to solving

– 2*x* – 4 ≥ 9 or – 2*x* – 4 ≤ – 9

which gives: – 2*x* ≥ 13 or – 2*x* ≤ – 5

⇒ *x* ≤ – 13 / 2 or *x* ≥ 5 / 2

The above solution set is written in interval form as follows:

] – ∞ , – 13 / 2] ∪ [5 / 2 , + ∞ [

and the graph of solution is sketched below.

*1*. Functions and their graph. One–one and onto functions. Composite and inverse functions.

*2*. Linear inequalities. Systems of linear inequalities

*3*. Polynomial inequalities. Rational inequalities. Absolute-value inequalities

*6*. Differentiation. Derivative rules, the chain rule.

*7*. Application of differentiation: L'Hospital's Rule

*8*. Vertical, Horizontal and Slant asymptotes

*9*. Higher Order Derivatives. Applications of differentiation: local and absolute extremes of a function

*10*. Curve Sketching

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