The discussion revolves around finding the composition of functions G(F(x)), where F(x) = √(2x-1) and G(x) = x/(x-2). Participants are tasked with determining the domain and range of this composition.
There is an ongoing examination of the conditions under which the functions are defined. Some participants are questioning the correctness of the domain values proposed, particularly x ≠ 3/2 and x ≤ 1/2, while others are exploring the implications of these conditions on the range.
Participants are navigating through the implications of square root functions and rational expressions, with specific attention to the constraints these impose on the values of x. There is a noted confusion regarding the algebraic manipulations leading to the domain restrictions.
If x < 1/2 , then you are taking then you are taking the square root of a negative number.Wa1337 said:Homework Statement
If F(x)=√(2x-1) and G(x) = x/(x-2), find G(F(x)) and its domain & range.
The Attempt at a Solution
G(F(x)) = ((√2x-1)/(√(2x-1)-2)
x ≠ 3/2, x </= 1/2
Where do I go from here?
SammyS said:If x < 1/2 , then you are taking then you are taking the square root of a negative number.
Perhaps you meant x ≰ 1/2 . If that's the case, it's more straight forward to say, x > 1/2 .
Your answer of x ≠ 3/2 is incorrect.
How about the range ?
√(4) = 2, so 2x - 1 = 4 will give you 2 - 2 which would be division by zero. That solution is not x = 3/2. Check your algebra.Wa1337 said:Well i thought the denominator could not equal 0, so I did √(2x-1) - 2 ≠ 0 and got x ≠ 3/2.
For the numerator I thought that square roots can't be negative, so they should be >/= 0. So i did √(2x-1) there and got x </= 1/2.
I don't know how to get the range from here, little confused