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# LetFind the range of . Give your answer as an interval.

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Nov 6, 2014

#1
+109518
+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)

https://www.desmos.com/calculator/p9jhxdm5ff

Nov 7, 2014

#1
+109518
+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)

https://www.desmos.com/calculator/p9jhxdm5ff

Melody Nov 7, 2014