solving inequalities in interval notation
- Thread starter kpx001
- Start date
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
No that will not work.
If (x+6) (x-2 ) > 0, then both factors have the same sign.
Thus, if x<-6 both are negative or if x>2 both are positive.
So the solution set is \(\displaystyle \left( { - \infty , - 6} \right) \cup \left( {2,\infty } \right).\)
If (x+6) (x-2 ) > 0, then both factors have the same sign.
Thus, if x<-6 both are negative or if x>2 both are positive.
So the solution set is \(\displaystyle \left( { - \infty , - 6} \right) \cup \left( {2,\infty } \right).\)
D
Deleted member 4993
Guest
kpx001 said:
Plot those functions in your graphing calculator and see how you can interpret pka's answer.
what about
(3 + x) / (3-x) >= 1
3+x >= 3 - x
i try combining but i dont get the right answer because 3 cant be it and i can guess and check [0,3). can anyone give me an explanation to solve this problem? i tihnk i need to split the problems into two inequalities but i dont know how to.
(3 + x) / (3-x) >= 1
3+x >= 3 - x
i try combining but i dont get the right answer because 3 cant be it and i can guess and check [0,3). can anyone give me an explanation to solve this problem? i tihnk i need to split the problems into two inequalities but i dont know how to.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Always make such problems comparable to zero:
\(\displaystyle \L \frac{{3 + x}}{{3 - x}} \ge 1\quad \Rightarrow \quad \frac{{3 + x}}{{3 - x}} - 1 \ge 0\).
So \(\displaystyle \L \frac{{\left( {3 + x} \right) - \left( {3 - x} \right)}}{{3 - x}} \ge 0\quad \Rightarrow \quad \frac{{2x}}{{3 - x}} \ge 0\).
Now we need the numerator and the denominator to have the same sign so that their quotient is not negative. Thus we need \(\displaystyle \L 2x \ge 0\quad \& \quad 3 - x > 0\) OR \(\displaystyle \L 2x \le 0\quad \& \quad 3 - x \le 0\)
\(\displaystyle \L \frac{{3 + x}}{{3 - x}} \ge 1\quad \Rightarrow \quad \frac{{3 + x}}{{3 - x}} - 1 \ge 0\).
So \(\displaystyle \L \frac{{\left( {3 + x} \right) - \left( {3 - x} \right)}}{{3 - x}} \ge 0\quad \Rightarrow \quad \frac{{2x}}{{3 - x}} \ge 0\).
Now we need the numerator and the denominator to have the same sign so that their quotient is not negative. Thus we need \(\displaystyle \L 2x \ge 0\quad \& \quad 3 - x > 0\) OR \(\displaystyle \L 2x \le 0\quad \& \quad 3 - x \le 0\)