SOLUTION: The one to one function f is defined on the domain x>0 by f(x) = (2x-1)/(x+2). A) State the range, A, of f. B) Obtain an expression for f^(-1) (x) for x∈A. I am assuming

Question 978200: The one to one function f is defined on the domain x>0 by f(x) = (2x-1)/(x+2).
A) State the range, A, of f.
B) Obtain an expression for f^(-1) (x) for x∈A.
I am assuming that the letter A is only a variable used to represent the range of f(x). If that is the case, how can an expression be made for the inverse function?
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!

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(wants more help)

That expression is essentially switching the places or roles of x and y. You want a function that will give you x.

Try not using the "...to the negative one power" type notation. You want a function g(x) so that f(x) and g(x) will undo each other. That means that .

You have your specific definition for f(x). You want to use some other function g(x) AS INPUT to f(x) and the OUTPUT must be x.

.

Form that last-specified composition:

Solve that equation for g(x).
The inverse of f(x) will be g(x).
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SOLUTION: The one to one function f is defined on the domain x>0 by f(x) = (2x-1)/(x+2). A) State the range, A, of f. B) Obtain an expression for f^(-1) (x) for x&#8712;A. I am assuming

SOLUTION: The one to one function f is defined on the domain x>0 by f(x) = (2x-1)/(x+2). A) State the range, A, of f. B) Obtain an expression for f^(-1) (x) for x∈A. I am assuming