Question 113011
{{{f(x)=(1)/(x+2)}}} Start with the given function



{{{x+2=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.




{{{x=0-2}}}Subtract 2 from both sides



{{{x=-2}}} Combine like terms on the right side






Since {{{x=-2}}} makes the denominator equal to zero, this means we must exclude {{{x=-2}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq-2\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>-2}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -2\right)\cup\left(-2,\infty \right)]



Now to find the range, notice that there isn't an x term in the numerator. So if 
{{{(1)/(x+2)=0}}}, there isn't an x value that will satisfy the equation. So in other words, {{{f(x)=(1)/(x+2)}}} will never be equal to zero. So this means we must take 0 out of our range.




So our range is:  *[Tex \LARGE \textrm{\left{y|y\in\mathbb{R} y\neq0\right}}]


which in plain English reads: y is the set of all real numbers except {{{y<>0}}}


So our range looks like this in interval notation

*[Tex \Large \left(-\infty, 0\right)\cup\left(0,\infty \right)]



Now let's graph {{{f(x)=(1)/(x+2)}}} 


{{{ graph( 500, 500, -10, 10, -10, 10, (1)/(x+2)) }}}



So we can see that at x=-2 there is a gap (the vertical piece is not part of the graph) and at y=0 there is an asymptote. So this verifies our answer.