1. ## (simple) Inequalities.

0 <= e^(-x) < 1

I have to find the solution set so i got:
minus infinity > x > 0
correct?

and for
|2-3x|<3
i get
x > 1/3
and
x < 5/3

now my question is:
on the second inequality, the solution set is the INTERSECTION of both solutions? (is it an OR or and AND connection of both sets?)

2. ## Re: (simple) Inequalities.

Originally Posted by nappysnake
0 <= e^(-x) < 1

I have to find the solution set so i got:
minus infinity > x > 0
correct?

and for
|2-3x|<3
i get
x > 1/3
and
x < 5/3

now my question is:
on the second inequality, the solution set is the INTERSECTION of both solutions? (is it an OR or and AND connection of both sets?)
$\displaystyle 1.$

$\displaystyle e^x > 0 \land e^x>1$

$\displaystyle 2.$

$\displaystyle |2-3x| =\begin{cases}3x-2, & \text{if }x>\frac{2}{3} \\2-3x, & \text{if }x \leq \frac{2}{3}\end{cases}$

$\displaystyle 2a.$

$\displaystyle x \in (\frac{2}{3},+\infty)$

$\displaystyle 3x-2<3$

$\displaystyle 2b.$

$\displaystyle x \in \left(-\infty,\frac{2}{3}\right]$

$\displaystyle 2-3x<3$

SolutionOf(2)=SolutionOf(2a) $\displaystyle \lor$ SolutionOf(2b)

3. ## Re: (simple) Inequalities.

For the second, I think it's really better to do it simultaneously. So

$\displaystyle |2-3x| < 3$

or

$\displaystyle -3 < 2-3x < 3$.

So $\displaystyle -5 < -3x < 1$

then $\displaystyle 5 > 3x > 1$

then $\displaystyle \frac{5}{3} > x > \frac{1}{3}$

or

$\displaystyle \frac{1}{3} < x <\frac{5}{3}$