SOLUTION: Can someone let me know what is
1+x divided by 1-x greater than equal to to 1
And i need something clearified. When solving absolute values of inequalities when plotting th
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Question 731903: Can someone let me know what is
1+x divided by 1-x greater than equal to to 1
And i need something clearified. When solving absolute values of inequalities when plotting the graph when do u shade in the circle on the graph and when do u plot above the line with th arrows instead of on the line.Thanks!!
Found 2 solutions by stanbon, josgarithmetic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Can someone let me know what is
1+x divided by 1-x greater than equal to to 1
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(1+x)/(1-x) >= 1
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(1+x) >= 1-x
2x >= 0
x >= 0
===============================
And i need something clearified. When solving absolute values of inequalities when plotting the graph when do u shade in the circle on the graph and when do u plot above the line with th arrows instead of on the line.
---
If x > 4 the point x = 4 is circled but not shaded,
Then an arrow is drawn to the right from x = 4.
----
If x >=4 the x = 4 is circled and shaded.
Then an arraw is drawn to the right from x = 4.
=============
Cheers,
Stan H.
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
1+x divided by 1-x greater than equal to to 1....
, you must permit the possiblities of 1-x being negative and 1-x being negative and check each of these conditions separately. 1-x must NOT be zero so x must NOT equal 1. This value would be a critical point for x for checking a couple of intervals.
Letting 1-x>0, 1+x>=1*(1-x), and you can solve for x for this condition.
1+x>=1-x
1-1+x+x>=1-1-x+x
2x>=0
x>=0
Check this interval. x=4, (1+4)/(1-4)=-5/3>=1 FALSE.
Letting 1-x<0, 1+x<=1*(1-x), and you can then solve for x in this condition.
1+x<=1-x
1-1+x+x<=1-1-x+x
2x<=0
x<=0
Check this interval. x=-4, (1-4)/(1+4)=-3/5>=1 FALSE.
What about the interval, 0<=x<1?
Try x=(1/2).
(1+1/2)/(1-1/2)=(3/2)/(1/2)=3>=1? TRUE.
I must have made a mistake which I have not been able to find, but using the critical points of x at 0 and 1 and the three intervals around them seems to work well. The solution is .
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