Question 138671
{{{x/(x+3)<0}}} Start with the given inequality


{{{x+3=0}}} Set the denominator equal to zero



{{{x=0-3}}}Subtract 3 from both sides



{{{x=-3}}} Combine like terms on the right side



So the vertical asymptote is {{{x=-3}}}. This is one critical value



{{{y=x/(x+3)}}} Start with the given equation


{{{0=x/(x+3)}}} Plug in {{{y=0}}}


{{{x=0}}} Solve for x



So our critical values are {{{-3}}} and {{{x=0}}}




Now set up a number line and plot the critical values on the number line


{{{number_line( 600, -10, 10,0,-3)}}}






So let's pick some test points that are near the critical values and evaluate them.



Let's pick a test value that is less than {{{-3}}} (notice how it's to the left of the leftmost endpoint):


So let's pick {{{x=-4}}}


{{{x/(x+3)<0}}} Start with the given inequality



{{{(-4)/(-4+3)< 0}}} Plug in {{{x=-4}}}



{{{4< 0}}} Evaluate and simplify the left side


Since the inequality is false, this means that the interval does <b>not</b> work. So this interval is <b>not</b> in our solution set and we can ignore it.




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Let's pick a test value that is in between {{{-3}}} and {{{0}}}:


So let's pick {{{x=-1}}}


{{{x/(x+3)<0}}} Start with the given inequality



{{{(-1)/(-1+3)< 0}}} Plug in {{{x=-1}}}



{{{-1/2< 0}}} Evaluate and simplify the left side


Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.

   So part our solution in interval notation is <font size="8">(</font>*[Tex \LARGE -3,0]<font size="8">)</font>



   



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Let's pick a test value that is greater than {{{0}}} (notice how it's to the right of the rightmost endpoint):


So let's pick {{{x=1}}}


{{{x/(x+3)<0}}} Start with the given inequality



{{{1/(1+3)< 0}}} Plug in {{{x=1}}}



{{{1/4< 0}}} Evaluate and simplify the left side


Since the inequality is false, this means that the interval does <b>not</b> work. So this interval is <b>not</b> in our solution set and we can ignore it.



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Summary:


So the solution in interval notation is:



<font size="8">(</font>*[Tex \LARGE -3,0]<font size="8">)</font>




And the graph of this solution set looks like this



{{{drawing(500,80,-13, 7,-10, 10,
number_line( 500, -13, 7),

line(-3,0,0,0),
line(-3,.3,0,.3),
line(-3,.15,0,.15),
line(-3,-.3,0,-.3),
line(-3,-.15,0,-.15),


circle(0,0,0.3),
circle(0,0,0.3),
circle(0,0,0.3),
circle(0,0,0.3-0.02),

circle(-3,0,0.3),
circle(-3,0,0.3),
circle(-3,0,0.3),
circle(-3,0,0.3-0.02)
)}}}