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Math Help - Fidn the domain of..

  1. #1
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    Find the domain of..

    Sqrt[ (1-x^2) / (ln(4-2x) ]

    Okay.. So, since it is a square root, I know everything under the radical must be >= 0, so

    Sqrt[ (1-x^2) / (ln(4-2x) ] >= 0

    But, since it is a rational fraction, the demoninator can't be zero.

    ln(4-2x) != 0

    I also know since it is a LN, the input must be > 0.. so..

    4-2x >= 0

    Am I on the right start here, in discovering the domain?

    4 - 2x >= 0
    x < 2


    ln(4-2x) != 0
    4-2x != e^0
    -2x != -4 + 1
    -2x != -3
    x != 3/2


    this just leaves hte last one, i think.. can someone help me?

    Thank you
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  2. #2
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    Quote Originally Posted by zodiacbrave View Post
    Sqrt[ (1-x^2) / (ln(4-2x) ]

    Okay.. So, since it is a square root, I know everything under the radical must be >= 0, so

    Sqrt[ (1-x^2) / (ln(4-2x) ] >= 0

    But, since it is a rational fraction, the demoninator can't be zero.

    ln(4-2x) != 0

    I also know since it is a LN, the input must be > 0.. so..

    4-2x >= 0

    Am I on the right start here, in discovering the domain?

    4 - 2x >= 0
    x < 2


    ln(4-2x) != 0
    4-2x != e^0
    -2x != -4 + 1
    -2x != -3
    x != 3/2


    this just leaves hte last one, i think.. can someone help me?

    Thank you
    f(x) = \sqrt{\frac{1 - x^2}{\ln{(4 - 2x)}}}.

    First, look for any restrictions that you know of.

    You know that you can not have the logarithm of a nonpositive number, so to start with

    4 - 2x > 0

    4 > 2x

    x < 2.

    Also, since you have a square root, everything under the square root must be nonnegative.

    So \frac{1 - x^2}{\ln{(4 - 2x)}} \geq 0.


    Note that since the denominator is ALWAYS positive, the numerator can not be negative.

    So 1 - x^2 \geq 0

    1 \geq x^2

    x^2 \leq 1

    |x| \leq 1

    -1 \leq x \leq 1.



    So we have in our domain

    x < 2 AND -1 \leq x \leq 1.

    For both inequalities to hold, -1 \leq x \leq 1.


    Thus the domain of f is -1 \leq x \leq 1.
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  3. #3
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    \frac{3}{2} < x < 2 is also in the domain.

    \ln(4-2x) < 0 in that interval, as is (1-x^2) , making \frac{1-x^2}{\ln(4-2x)} > 0
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  4. #4
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    Quote Originally Posted by skeeter View Post
    \frac{3}{2} < x < 2 is also in the domain.

    \ln(4-2x) < 0 in that interval, as is (1-x^2) , making \frac{1-x^2}{\ln(4-2x)} > 0
    Edit - you are right...
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