Question 1121667
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.


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