One to one function h is defined by h(x)=(3x-4)/(-2x+9). Find h^1, the inverse of h. Then give the domain and range of h^1 using interval notation.
h(x)=(3x-4)/(-2x+9)
Inverse: h^-1(x)= (9x+4)/(2x+4)

Don't know how to do the domain and range. Think domain is (-infinity, 9) union (4, infinity) and range is (negative infinity, infinity).

Been on this all day- can't find anything or anyone that makes sense... please help

Question

One to one function h is defined by h(x)=(3x-4)/(-2x+9). Find h^1, the inverse of h. Then give the domain and range of h^1 using interval notation.  
h(x)=(3x-4)/(-2x+9)
Inverse: h^-1(x)= (9x+4)/(2x+4) 

Don't know how to do the domain and range. Think domain is (-infinity, 9) union (4, infinity) and range is (negative infinity, infinity).  

Been on this all day- can't find anything or anyone that makes sense... please help

anonymous · Answer

Inverse: h^-1(x)= (9x+4)/(2x+4) 

Don't know how to do the domain and range. Think domain is (-infinity, 9) union (4, infinity) and range is (negative infinity, infinity).  

Been on this all day- can't find anything or anyone that makes sense... please help

anonymous · Answer

domain of $h$ is all real numbers except $\frac{9}{2}$ where the denominator is zero

anonymous · Answer

inverse is either a mistake or a typo
it is 
$$h^{-1}(x)=\frac{9x+4}{2x+3}$$

anonymous · Answer

domain of $h^{-1}$ is all real numbers except $-\frac{3}{2}$ which is not surprising , because the range of $h$ is all real numbers except $-\frac{3}{2}$

anonymous · Answer

It has to be in interval notation. That's my problem- it's an equations where it's one to one. I've inversed the original equation, but can't find out how to put it into interval form for the domain and range. 

I would name my first born satellite73 if you get this one ;)

anonymous · Answer

original was h(x) = 3x-4
                            ______
                           -2x+9

anonymous · Answer

interval notation for all real numbers except $\frac{9}{2}$ is the silly looking 
$$(-\infty, \frac{9}{2})\cup (\frac{9}{2},\infty)$$

anonymous · Answer

all you need to do for the domain is to set the denominator equal to zero and solve for $x$
then say "all numbers but that one"

anonymous · Answer

OMG you're my hero!

anonymous · Answer

somehow you made a tiny mistake on the inverse, it is not what you wrote, it is the one i wrote above

anonymous · Answer

and clearly $\frac{3x-4}{-2x+9}$ cannot be $-\frac{3}{2}$ 
if you look at it carefully you will see why

anonymous · Answer

yes i did the 4 was supposed to be a 3 in the denominator

And the range

anonymous · Answer

domain of $h$ is the range of $h^{-1}$ and vice versa

anonymous · Answer

gotta run, good luck

anonymous · Answer

thank you!