#### 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 $\frac{1}{4-2x}<3$. The domain of the inequality is $D=\mathbb{R}\setminus \{2\}$ since only for these values of $x$ the denominator is non-zero. If the inequality is multiplied by the term $4-2x$, three cases have to be distinguished. Fill in the blanks in the following text accordingly:

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

- On the interval

the term is positive, the comparator remains unchanged, and the new inequality reads $1\hspace{0.5em}<\hspace{0.5em}$

. Linear transformations result in the solution set $L{}_{1}$$\hspace{0.5em}=\hspace{0.5em}$

. 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}$$\hspace{0.5em}=\hspace{0.5em}$

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.

##### **Exercise 3.2.5 **

The solution set of the inequality $\frac{x-1}{x-2}\le 1$ is $L$$\hspace{0.5em}=\hspace{0.5em}$ .

##### **Exercise 3.2.6 **

The solution set of the inequality $\frac{1}{1-\sqrt{x}}<1+\sqrt{x}$ is $L$$\hspace{0.5em}=\hspace{0.5em}$

.

.