Question 1088255
<font color="black" face="times" size="3">The function {{{f(x) = (3x-7)/(x+1)}}} has the horizontal asymptote {{{(3x)/x = 3/1 = 3}}}. 


You divide the leading terms where {{{x <> 0}}}. 


As x gets larger and larger, f(x) will approach f(x) = 3.


What this means is that any value but y = 3 is possible for the range. 


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If we have {{{y = (3x-7)/(x+1)}}} and we swapped x and y, then we'd have {{{x = (3y-7)/(y+1)}}}


Let's solve for y


{{{x = (3y-7)/(y+1)}}}


{{{x(y+1) = 3y-7}}}


{{{xy+x = 3y-7}}}


{{{xy-3y = -7-x}}}


{{{y(x-3) = -7-x}}}


{{{y = (-7-x)/(x-3)}}}


which is the inverse function. The denominator of this inverse, x-3, indicates that 3 is not in the domain of the inverse. This confirms that 3 is not in the range of the original function.


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So the range is the set of all y values such that y is a real number but y cannot equal 3. 


In set builder notation, we'd say *[Tex \Large \left\{y|y\in \mathbb{R}, \ y \neq 3\right\}]


In interval notation, we'd write *[Tex \Large \left(-\infty, 3\right) \cup \left(3, \infty\right)]


The big U means "union" to essentially "glue" the two intervals together.


Here's what the graph looks like. The blue dashed line is the horizontal asymptote y = 3 (it goes through the two points (0,3) and (1,3))
<img src = "https://image.prntscr.com/image/LyUtwAtmQxuBN03e6Gnl6A.png">
The green curve approaches the dashed line but it will never cross it or touch it. So this shows us that any output but y = 3 is possible. 

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