Question

find the range of f(x)= x+1 / x-2

Answer 1

For starters, the function has a vertical asymptote at \(x=2\). At that point, you should at least see the range go to positive and negative infinity. However, the range isn't \(\large (-\infty,\infty)\)! Can you see a potential value(s) that might not be in the range of your function? :-)

Answer 2

is it (-infinity,+infinity) except 2

Answer 3

That would be it's domain (all x values except where he have vertical asymptotes, holes, etc.). The range is \(\large (-\infty,\infty)\) except some y value. To get a better idea for what this y value is, what do you get for the horizontal asymptotes of the function \(\large f(x) = \dfrac{x+1}{x-2}\)?

Answer 4

is it except -1 & 2?

Answer 5

No; 2 is a vertical asymptote and doesn't tell you what value to exclude from the interval \(\large (-\infty,\infty)\). The role of the vertical asymptote was to help us see if the function is approaching \(\large \infty,\,-\infty\) or both. In the case of \(\large f(x)=\dfrac{x+1}{x-2}\), we have the function approaching both infinities. Thus, the only thing we need to check for in the range is whether or not there are horizontal asymptotes (since the function can never equal the values of HA's and thus aren't included in the range of our function). In this case, you correctly computed that the only horizontal asymptote of this function is \(\large y=1\). Therefore, we conclude that the range is \(\large (-\infty,\infty)\) except 1; otherwise written as \(\large (-\infty,1)\cup(1\infty)\).

I hope this makes sense! :-)

Answer 6

That last interval should read \(\large (-\infty,1)\cup (1,\infty)\) (forgot the comma). Sorry about that! :-/

Answer 7

:) yea... i undrstand alrdy...:) thnx :)