|
Question 1121667: I don't know how to start this.
2/x+3 - 1/x+1 < 0
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20850)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 2/(x+3) - 1/(x+1) < 0
this function is undefined when x = -3 and when x = -1.
this function is equal to 0 when x = 1.
to solve for that, set the function equal to 0.
you get 2/(x+3) - 1/(x+1) = 0
multiply both sides of the equiation by (x+3) * (x+1) to get:
2 * (x+1) - (x + 3) = 0
simplify to get 2x + 2 - x - 3 = 0
combine like terms to get x - 1 = 0
solve for x to get x = 1
you have 4 intervals that need to be checked.
they are:
x < -3
x between -3 and -1
x between -1 and 1
x > 1
the function is continuous between these intervals.
therefore, within each interval, if it is greater than 0, it stays greater than 0 and, if it is less than 0, it says less than 0.
it can only go from positive to negative at either side of the vertical asymptotes or when x = 1.
the function is 2/(x+3) - 1/(x+1) < 0
when x = -4, the function is 2/-1 - 1/-3 = -2 + 1/3 = less than 0.
when x = -2, the function is 2/1 - 1/-1 = 2 + 1 = greater than 0.
when x = 0, the function is 2/3 - 1 = less than 0.
when x = 2, the function is 2/5 - 1/3 = greater than 0.
it appears that the function is less than 0 when x < -3 and when x is between -1 and 1.
the following graph confirms that.
|
|
|
| |
