#1**+5 **

$$f(x)=\frac{3x-7}{x+1}$$

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)

Melody
Nov 7, 2014

#1**+5 **

Best Answer

$$f(x)=\frac{3x-7}{x+1}$$

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)

Melody
Nov 7, 2014