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Math Help - Functions

  1. #1
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    Functions

    The intersection of the domains of functions: f(x) = arcsin(log\frac{x}{10}) and g(x) = \sqrt{2+x-x^2}

    a) [1,2]
    b) (0,2]
    c) [1, \infty)
    d) \phi
    e) \mathbb{R}_+^*
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    The intersection of the domains of functions: f(x) = arcsin(log\frac{x}{10}) and g(x) = \sqrt{2+x-x^2}

    a) [1,2]
    b) (0,2]
    c) [1, \infty)
    d) \phi
    e) \mathbb{R}_+^*
    1. g(x) = \sqrt{2+x-x^2}

    Solve for x:

    2+x-x^2 \geq 0

    I've got x \in [-1, 2]

    2. The domain of the arcsin-function is [-1, 1].

    I assume that \log means \log_{10}

    \log\left(\dfrac x{10}\right) = -1~\implies~x = 1

    \log\left(\dfrac x{10}\right) = 1~\implies~x = 100

    That means the domain of the given arcsin-function is [1, 100]

    3. Now calculate the intersection. (I've got answer a))
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  3. #3
    Senior Member Twig's Avatar
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    hi

    Easiest is looking at the second function maybe.
    For the expression under the square root sign to be positivt, we solve
     -x^2 + x +2 = (x-2)(x+1)
    So, this function makes boundaries  [-1,2]

    Looking at  arcsin(log(\frac{x}{10})) = arcsin(log(x) - log(10))

    That is,  arcsin(log(x) - 1)

    The logharitm is defined for  x>0 , so x>0
    and arcsin(x) for  |x| \leq 1 .

    So we get  |log(x) - 1| \leq 1 , giving us
    when we combine the interval  [1,2]
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  4. #4
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    thank you
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