How do I solve this nonlinear inequality?

Question

1/x+3 + 1/x+4 is less than or equal to 0

 

Picture of the problem

 

http://i.imgur.com/pvojkPT.png

 

Edit: The answer is (-infinity, -4) U [-7/2, -3)

 

How do I get that answer?

Answer 1

1/(x+3) + 1/(x+4) ≤ 0

As Michael did, we need to get the common denominator of (x+3)(x+4)

This worked out to (2x+7)/((x+3)(x+4)) ≤ 0

 

The zero occurs at 2x + 7 = 0 so the zero is at x= -3.5
There is a vertical asymptote at x=-3 and x=-4

There are several areas we need to test

1)  x<-4

2)  -4<x<-3.5

3)  -3.5<x<-3

4)  x>-3
 
try x= -5... we get y=-3/2 or -1.5 which is ≤ 0 so this works
so numbers less than -4 are in the solution
 
try a number between the two vertical asymptotes
try x=-3.7 ... we get
    (2(-3.7)+7)/((-3.7+3)(-3.7+4))= 
    (-.4)/(-.7)(.3) = (-.4)/(-.21)=1.905 ≥0
so numbers between -3.5 and -4 do not work
 
Try a number between x=-3.5 and x=-3 (the other asymptote)
try x = -3.4
(2(-3.4)+7)/((-3.4+3)(-3.4+4))
= (.2)/((-.4)(.6)) = (.2)/(-.24) = -.83333 ≤ so this does work so the area between x=-3.5 and x=-3 is in the solution.
 
Now try a number to the right of our x=-3 asymptote.
Try x=-1

(2(-1)+7)/((-1+3)(-1+4)) = 5/6 ≥ 0 so the numbers to the
right of x=-3 are not in the solution.
 
To summarize:

 

x <-4 are included
-4<x<-3.5 are not included
x=-3.5 is included
-3.5<x<-3 are included
x>-3 is not included

 

Final answer (-∞<x<-4) ∪ (-3.5≤x<-3)
 
In interval notation   (-∞,-4)∪[-3.5,-3) 
 

Answer 2

[1 / (x + 3)] + [1 / (x + 4)] ≤ 0

 

 

We have a rational inequality, which means that we need to use the LCD.  The LCD is (x + 3)(x + 4).

 

 

((x + 4) + (x + 3)) / (x + 3)(x + 4) ≤ 0

 

(2x + 7) / (x + 3)(x + 4) ≤ 0

 

 

Set the numerator of the inequality equal to zero.

 

2x + 7 = 0

 

2x = -7

 x = -3.5

 

 

Okay.   Since the inequality is equal to or less than 0, we need to find values of x so that when the inequality is evaluated, the left side is 0 or negative.  In a graph, the values of y would be on or below the x-axis.

 

x = -3.5 is our zero.  x = -4   and x = -3  are our vertical asymptotes.   we will use test points around these x values to evaluate f(x).

 

Lets use test points:  x=-5  ,  x=-3.7 ,  x=-3.3  ,  x=-2

 

f(-5) = (2(-5) + 7) / (-5 + 3)(-5 + 4)

        = -3 / (-2 * -1)

        = -3/2

 

f(-3.7) = (2(-3.6) + 7) / (-3.5 + 3)(-3.5 + 4)

           = -0.2 / (-0.24)

           = 0.83

 

f(-3.3) = (2(-3.3) +7) / (-3.3 + 3)(-3.3 + 4)

          = 0.4 / (-0.21)

          = -1.90

 

f(-2) = (2(-2) + 7) / (-2 + 3)(-2 + 4)

       = 3 / (1 * 2)

       = 3/2

 

The intervals in which f(x) is zero or negative is

 

x = (∞ , -4)∪(-4 , -3)