What Is the Domain and Range of G(F(x))?

[Wa1337]

Homework Statement

If F(x)=√(2x-1) and G(x) = x/(x-2), find G(F(x)) and its domain & range.

Homework Equations

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?

 

[flyingpig]

x \leq \frac{1}{2} is wrong

How did you get x ≠ 3/2?

 

[SammyS]

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?

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 ?

 

[Wa1337]

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 ?

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

 

[SammyS]

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

√(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.

You said "For the numerator I thought that square roots can't be negative, so ... ". That's not quite right. What is true for this case is that you can't take the square root of a negative number. Therefore, you need 2x-1 ≥ 0 . That doesn't give you x ≤ 1/2 .