Chapter 3 Inequalities in one Variable

Section 3.2 Transformation of Inequalities

3.2.2 Exercises

If the inequality is multiplied by a composite term, we must investigate precisely for which values of

x

the case analysis must be done:

Exercise 3.2.4

Find the solution set of the inequality

\frac{1}{4 - 2 x} < 3

. The domain of the inequality is

D = ℝ ∖ {2}

since only for these values of

x

the denominator is non-zero. If the inequality is multiplied by the term

4 - 2 x

, three cases have to be distinguished. Fill in the blanks in the following text accordingly:

On the interval
the term is positive, the comparator remains unchanged, and the new inequality reads $1 <$
. Linear transformations result in the solution set $L_{1}$ $=$
. The elements of this set satisfy the case condition.
On the interval
the term is negative, the comparator is inverted. Initially, the new inequality has the solution set
, because of the case condition only the subset $L_{2}$ $=$
is allowed.
The single value $x = 2$ is no solution of the initial inequality since it is not in .

Sketch the solution set of the inequality and indicate the boundary points.

On the interval

] - \infty; 2 [

the term is positive. The corresponding solution set is

] - \infty; \frac{11}{6} [

. In contrast, on the interval

] 2; \infty [

the term is negative, the comparator is inverted. Initially, the new inequality has the solution set

] \frac{11}{6}; \infty [

, because of the case condition

x > 2

only the subset

L_{2} =] 2; \infty [

is allowed. So, altogether the union set

L = L_{1} \cup L_{2} = ℝ ∖ [\frac{11}{6}; 2]

is the solution set of the initial inequality excluding the boundary points:

Exercise 3.2.5

The solution set of the inequality

\frac{x - 1}{x - 2} \leq 1

L

=

The domain of the inequality is

D = ℝ ∖ {2}

For $x > 2$ , multiplying the inequality by the term $x - 2$ results in $x - 1 \leq x - 2$ , which is equivalent to the false statement $- 1 \leq - 2$ . Thus, this case does not contribute a solution to the solution set.
For $x < 2$ , multiplying the inequality by the term $x - 2$ results in $x - 1 \geq x - 2$ , which is equivalent to the true statement $- 1 \geq - 2$ . Because of the case condition the solution interval for this case is only $L_{2} =] - \infty; 2 [$ .
The single value $x = 2$ is no solution.

So, altogether the solution set is

L =] - \infty; 2 [

excluding the boundary points (even though the comparator

\leq

occurred in the initial inequality).

Exercise 3.2.6

The solution set of the inequality

\frac{1}{1 - \sqrt{x}} < 1 + \sqrt{x}

L

=

The domain of the inequality is

D = [0; \infty [∖ {1}

since only for these values of

x

the square root is defined and the denominator is non-zero.

For $0 \leq x < 1$ , multiplying the inequality by the term $1 - \sqrt{x}$ results in $1 < (1 + \sqrt{x}) (1 - \sqrt{x})$ , which is equivalent to $1 < 1 - x$ . This inequality is satisfied for $x < 0$ , but these values of $x$ violate the case condition and thus, they are not in the solution set.
For $x > 1$ , multiplying the inequality by the term $1 - \sqrt{x}$ results in $1 > 1 - x$ , which is equivalent to $x > 0$ . But only the values of $x$ in the interval $] 1; \infty [$ satisfy the case condition, hence $L =] 1; \infty [$ is the only solution interval of the initial inequality.
The single value $x = 1$ is no solution.

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics

Chapter 3 Inequalities in one Variable

3.2.2 Exercises

Exercise 3.2.4

Exercise 3.2.5

Exercise 3.2.6