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Homework Help: Let f, g and h be functions defined below:

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

    which of the following are true:
    A)Df={x∈ℝ:x≤1 or x≥2}

    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)

    g(x)=3/(√(-3+3)) gets 3/0 Undefined?
    g(x)=3/(√(-2+3)) (yes this is ok)

    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,

    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


    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
    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


    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
    if x is 2 or 3 it will be zero,
    but how can 0/0 undefined = ℝ?
  14. Jun 5, 2015 #13


    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


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    Staff Emeritus
    Science Advisor
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    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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