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Let f, g and h be functions defined below:

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    f(x)=(√x^2-3x+2)/(2x-3),
    g(x)=3/(√(x+3)) and
    h(x)=(x^2-5x+6)/(x-2)

    which of the following are true:
    A)Df={x∈ℝ:x≤1 or x≥2}
    B)Dg={x∈ℝ:x≥-3}
    C)Dh=ℝ

    2. Relevant equations


    3. The attempt at a solution
    I am using substitution here by replacing the x by the parameters specified and slightly further:

    Df={x∈ℝ:x≤1 or x≥2}
    f(x)=(√(0^2-3(0)+2))/(2(0)-3) (yes this is ok)
    f(x)=(√(1^2-3(1)+2))/(2(1)-3) (yes this is ok)
    f(x)=(√(-1^2-3(-1)+2))/(2(-1)-3) (yes this is ok)
    f(x)=(√(2^2-3(2)+2))/(2(2)-3) (yes this is ok)
    f(x)=(√(3^2-3(3)+2))/(232)-3) (yes this is ok)

    Dg={x∈ℝ:x≥-3}
    g(x)=3/(√(-3+3)) gets 3/0 Undefined?
    g(x)=3/(√(-2+3)) (yes this is ok)

    Dh=ℝ
    h(x)=(x^2-5x+6)/(x-2)
    h(x)=(1^2-5(1)+6)/(1-2) (yes this is ok)
    h(x)=(2^2-5(2)+6)/(2-2) (yes this is ok)

    So my answer would be:
    A and C are true.
     
  2. jcsd
  3. Jun 5, 2015 #2
    Your answers are correct.
    [EDIT:-however your reasoning for ##h(x)## is wrong. 2^2-2*5+6/(2-2)=0/0, which is indeterminate.
    HINT:-factorize the numerator of ##h(x)## to see why your answer is correct nonetheless.]
     
    Last edited: Jun 5, 2015
  4. Jun 5, 2015 #3
    0/0 is just 0 isn't it?
     
  5. Jun 5, 2015 #4
    They are the same.
     
  6. Jun 5, 2015 #5
    So why would you say in is undefined?
     
  7. Jun 5, 2015 #6
    No. 0/0 is indeterminate. The previous post was for a post that I think you deleted?
     
  8. Jun 5, 2015 #7
    So wouldn't only A be correct/true,

    Factorise
    h(x)=(2^2-5(2)+6)=(4-10+6)=0
    I don't think I understand...
     
  9. Jun 5, 2015 #8
    C is also correct.
    Factorize ##x^2-5x+6##, then see if anything changes in ##h(x)## because of that.
     
  10. Jun 5, 2015 #9

    Mark44

    Staff: Mentor

    No, the above should be h(1). h(x) = ##\frac{x^2 - 5x + 6}{x - 2}##
    This (above) should be h(2), which is NOT OK. As certainly said, this results in the indeterminate form 0/0, which should suggest that 2 is not in the domain of h.
     
  11. Jun 5, 2015 #10
    Mark,
    so is it only A?
    Certainly agrees with my initial answer, but I am not sure why?
    He has asked me to factor as above, I have but still not clear why?
     
  12. Jun 5, 2015 #11

    Mark44

    Staff: Mentor

    Write the formula for h(x) in factored form. IOW,
    $$h(x) = \frac{(?)(?)}{x - 2}$$

    In the factored form it is easier to see that for some value(s) of x, both the numerator and denominator are zero. This means that at any such value, the function h is not defined, so the domain is not all of R (the reals).
     
  13. Jun 5, 2015 #12
    (x-3)(x-2)
    if x is 2 or 3 it will be zero,
    but how can 0/0 undefined = ℝ?
     
  14. Jun 5, 2015 #13

    Mark44

    Staff: Mentor

    The full answer to my question is: ##h(x) = \frac{(x-3)(x-2)}{x - 2}##
    h(3) = 0/1 = 0
    h(2) is not zero. Since division by zero is not defined, 2 is not in the domain of h.

    0/0 is not a number -- it is one of several indeterminate forms. These usually show up as limits, which I'm pretty sure you haven't covered yet.
     
  15. Jun 5, 2015 #14
    I was a bit confused there about something, Mark is right, the domain is all of ##\mathbb{R}## except 2, so yeah, C is wrong.
     
  16. Jun 5, 2015 #15

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    When determining the domain here, you don't need to look at what the numerator is equal to. You simply have to note that the denominator is 0 when x=2, so h isn't defined for x=2. Division by 0 is always undefined.
     
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