Let
Find the range of 
. Give your answer as an interval.
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+118608 $$f(x)=\frac{3x-7}{x+1}$$
find the range.
I cannot remember doing questions like this so my method may not be the method taught.
I am feeling my way.
You cannot divide by 0 so x+1 cannot =0, x cannot equal -1
This is a restriction on the domain - not on the range.
x=-1 will be a vertical asymptote.
Just looking at it I can see it will be a hyperbola. (it reminds me of y=1/x)
We need the horizontal asymptote
ok
$$\\f(x)=\frac{3x-7}{x+1}\\\\
f(x)=\frac{3(x+1)-10}{x+1}\\\\
f(x)=\frac{3(x+1)}{x+1}+\frac{-10}{x+1}\\\\
f(x)=3+\frac{-10}{x+1}\\\\
$now $\;\; \frac{-10}{x+1} \;\;$ cannot equal zero $\\\\
so\;\;f(x)\;\;$cannot equal 3$\\\\$$
$$$The range of f is $\;\; (-\infty,3),(3,+\infty)\;\;
$I think that is in interval notation$\\\\
$ I think I would normally write is as $\;\;f(x)\in R\;\; where\;\; f(x)\ne3$$
Here is the graph (asymptotes are shown)
+118608 $$f(x)=\frac{3x-7}{x+1}$$
find the range.
I cannot remember doing questions like this so my method may not be the method taught.
I am feeling my way.
You cannot divide by 0 so x+1 cannot =0, x cannot equal -1
This is a restriction on the domain - not on the range.
x=-1 will be a vertical asymptote.
Just looking at it I can see it will be a hyperbola. (it reminds me of y=1/x)
We need the horizontal asymptote
ok
$$\\f(x)=\frac{3x-7}{x+1}\\\\
f(x)=\frac{3(x+1)-10}{x+1}\\\\
f(x)=\frac{3(x+1)}{x+1}+\frac{-10}{x+1}\\\\
f(x)=3+\frac{-10}{x+1}\\\\
$now $\;\; \frac{-10}{x+1} \;\;$ cannot equal zero $\\\\
so\;\;f(x)\;\;$cannot equal 3$\\\\$$
$$$The range of f is $\;\; (-\infty,3),(3,+\infty)\;\;
$I think that is in interval notation$\\\\
$ I think I would normally write is as $\;\;f(x)\in R\;\; where\;\; f(x)\ne3$$
Here is the graph (asymptotes are shown)
