Need some quick Math help (Rational Functions etc)

[Tbo]

Any help would be appreciated
I'm only looking help for possibly question 2 and 3.

So; f(x)=(2x-3)/(x-2)

1) Sketch the graph (easy enough, simple rational function)
2) Find the inverse and state the domain of the inverse
(When I found the inverse, it solved to be the exact same function (f(x)=f^-1(x)), i'm wondering if this is possible or if I've solved it wrong) f^-1 Domain= f(x) Range so ya
3) Find the values of x for which f(x)=f^-1(x)
This is the one that (sort of) has me stumped. Since they are the same function, do I write infinite solutions? Because it's a 2 mark question I'm assuming there's more to it.


Thanks for any help.

 

f(x)=(2x-3)/(x-2)
f^-1(x) = 1/f(x)

Therefore f^-1(x) = (x-2)/(2x-3) which therefore the domain and range are definitely not the same since the vertical asymptote is now different... went from x= 3/2 to now x=2.

As for when they're equal set f^-1(x) = f(x)
(x-2)/(2x-3) = (2x-3)/(x-2)
Isolated, distribute and solve.

0=(2x-3)^2/(x-2)^2 and then solve for values of x.

 

If you want me to write it out by hand and include a full solution let me know through pm or reply here. I thought I'd just get you started

 

[Tbo]

f(x)=(2x-3)/(x-2)
f^-1(x) = 1/f(x)

Therefore f^-1(x) = (x-2)/(2x-3) which therefore the domain and range are definitely not the same since the vertical asymptote is now different... went from x= 3/2 to now x=2.

As for when they're equal set f^-1(x) = f(x)
(x-2)/(2x-3) = (2x-3)/(x-2)
Isolated, distribute and solve.

0=(2x-3)^2/(x-2)^2 and then solve for values of x.

That would be the reciprocal, the inverse is when you swap y with x (in this case, x=(2y-3)/(y-2), solve for y. Sent you a pm though, ty.

 

Wasn't digging typing anything out so I wrote it.

http://imgur.com/qytNsZo.

You can try it yourself on any graphing calculator if you're really unsure