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'm sorry for t

Question 978092: 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'm sorry for the two part question, but I'm really confused as to what A is.
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
This "A" seems too much a mystery, and not common label for any typical set of numbers. The best you can do is to find the range of f(x); find the inverse, f^(-1)(x), and determine the range for this inverse.

The critical x values are x at 1/2 and at -2. f is undefined at x=-2, but domain is therefore all the rest of the real numbers. The RANGE for f^(-1) will be all real numbers such that .

Use the interval testing for range of f, but here is the graph to help:

The first function, f(x):

Notice both a horizontal and a vertical asymptotes.
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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'm sorry for t

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'm sorry for t