SOLUTION: The one to one function g is defined by g(x)= 5x/x-9 Find g-1 x and domain and range of g inverse x, using interval notation. I think inverse is 9x/x-5

Algebra ->  Inverses -> SOLUTION: The one to one function g is defined by g(x)= 5x/x-9 Find g-1 x and domain and range of g inverse x, using interval notation. I think inverse is 9x/x-5       Log On


   

Question 892069: The one to one function g is defined by g(x)= 5x/x-9
Find g-1 x and domain and range of g inverse x, using interval notation.
I think inverse is 9x/x-5
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Here is some help with the inverse part.

You likely are starting with g%28x%29=5x%2F%28x-9%29

You want to find g%5E%28-1%29%28x%29.
5g%5E%28-1%29%28x%29%2F%28g%5E%28-1%29%28x%29-9%29=x.
SOLVE THAT FOR g%5E%28-1%29%28x%29.

Let me use y instead of all that text...
5y%2F%28y-9%29=x
5y=x%28y-9%29
5y=xy-9x
5y-xy=-9x
y%285-x%29=-9x
y=-9x%2F%285-x%29
multiply by %28-1%29%2F%28-1%29
highlight%28y=9x%2F%28x-5%29%29-----the inverse

DOMAIN:
All real numbers except that x%3C%3E5.
(-infinity,5) U (5,infinity)

RANGE:
You figure this! What happens as x tends toward -infinity or toward infinity?
What happens to g%5E%28-1%29 near x=5, on either sides?