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G(F(x)) and domain/range

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    If F(x)=√(2x-1) and G(x) = x/(x-2), find G(F(x)) and its domain & range.

    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Oct 2, 2011 #2
    [tex]x \leq \frac{1}{2}[/tex] is wrong

    How did you get x ≠ 3/2?
     
  4. Oct 2, 2011 #3

    SammyS

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    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 ?
     
  5. Oct 2, 2011 #4
    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
     
  6. Oct 2, 2011 #5

    SammyS

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    √(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 .
     
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