Sketching reciprocal fn (1 Viewer)

[WEMG]

Hi everyone,
I was just doing inverse trigonometric fn exercises in the textbook, but decided not to skip the revision exercise(revisit of inverse fn's) included in it since I having touched it for a while.

There is a question which requires me to sketch g^-1(x) = (2-x)/(x-1). How do I know which quadrant the curves are in for any reciprocal fns?

 

[Trebla]

Reciprocal functions are not the same as inverse functions. Please clarify your question a bit more.

 

[iSplicer]

Hi everyone,
There is a question which requires me to sketch g^-1(x) = (2-x)/(x-1). How do I know which quadrant the curves are in for any reciprocal fns?

Test a point. Sub in x=1 and see what happens =]

 

[WEMG]

@Trebla
Sorry for not making my question clear enough, the question asks me to sketch the inverse of g(x) = (x+2)/(x+1) which is g^-1 = (2-x)/(x-1).

I thought that because the inverse function had a negative, I assumed the graph would have been in 2nd and 4th quadrant which is wrong..

@iSplicer

Thanks, I always thought that if there is a negative in a reciprocal, I can just flip it around like I do with parabolas.. haha

 

[hscishard]

huh? f(^-1)[x] means inverse and (f[x])^-1 means reciprocal.

-(x+2)/(x+1) is not (2-x)/(x-1)...