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How do you find the domain just by looking at the equation?

  1. Feb 23, 2005 #1
    How do you find the domain just by looking at the equation?

    ex1) x^2/ x^4+4x-5
    2) t/√(t^2-5t+6)
    3) √[2-√(4-x)]
     
  2. jcsd
  3. Feb 23, 2005 #2

    arildno

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    Take your last expression.
    What must you require of the argument of the (real) square root function?
     
  4. Feb 23, 2005 #3

    dextercioby

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    Take the first.What's the point in which the division (employed by the fraction) has no meaning...?

    Daniel.
     
  5. Feb 23, 2005 #4
    when x^4+4x-5=0
     
  6. Feb 23, 2005 #5

    arildno

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    gillgill:
    LEARN TO WRITE MATHS PROPERLY!! :grumpy:
    For 1) Did you mean x^2/(x^4+4x-5), x^2/(x^4+4x)-5, x^2/x^4 +4x-5
    Do you understand what I'm talking about?
     
  7. Feb 23, 2005 #6

    HallsofIvy

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    Assuming that your first function was x^2/ (x^4+4x-5) (do you see why I put in the parentheses?) then, yes, the denominator will be 0 when x^4+ 4x- 5= 0 and, since you cannot divide by 0, the domain is "all real numbers except solutions to x^4+ 4x- 5".

    In problem 2, the quantity inside the square root cannot be negative or zero (since we are dividing by it).

    In the last one, certainly we must have 4- x>= 0 (so x<= 4) in order to be able to do that square root but we must also have 2- sqrt(4-x)> 0. To determine what restriction that puts on x, look at 2- sqrt(4-x)= 0. That is the same as 2= sqrt(4-x) and, squaring, 4= 4-x which is the same as x= 0! If x< 0, then 4-x> 4,sqrt(4-x)> 2 and 2- sqrt(4-x)< 0. The domain is 0<= x<= 4.
     
  8. Feb 23, 2005 #7

    dextercioby

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    It's not the one with the 5 in the denominator,i hope,else he would have to solve that quartic...Actually a cubic,because "+1" is a sollution...

    Daniel.
     
  9. Feb 23, 2005 #8

    arildno

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    In addition to HallsofIvy's suggestions, I would like to say that what they are asking you to find, is the MAXIMAL domain of the (real) functions over the real numbers.
     
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