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Thread: 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
    $\displaystyle f(x) = \sqrt{\frac{1 - x^2}{\ln{(4 - 2x)}}}$.

    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

    $\displaystyle 4 - 2x > 0$

    $\displaystyle 4 > 2x$

    $\displaystyle x < 2$.

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

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


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

    So $\displaystyle 1 - x^2 \geq 0$

    $\displaystyle 1 \geq x^2$

    $\displaystyle x^2 \leq 1$

    $\displaystyle |x| \leq 1$

    $\displaystyle -1 \leq x \leq 1$.



    So we have in our domain

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

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


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

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

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