#### 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 $D=ℝ\setminus \left\{2\right\}$ 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:
1. 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}$$=$
. The elements of this set satisfy the case condition.
2. On the interval
the term is negative, the comparator is inverted. Initially, the new inequality has the solution set
, because of the case condition only the subset $L{}_{2}$$=$
is allowed.
3. 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$$=$ .

##### Exercise 3.2.6
The solution set of the inequality $\frac{1}{1-\sqrt{x}}<1+\sqrt{x}$ is $L$$=$
.