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Math Help - Finding domain of a function.

  1. #1
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    Finding domain of a function.



    Find the domain of f(x).

    Then find the inverse of the function, and the domain of the inverse.

    I know how to find the domain and inverse, but I just have no idea how to do it in this function.

    For domain, I'm guess 3 - e^2x > or = 0

    -e^2x > or = -3
    e^2x < or = 3
    2x < or = ln3
    x < or = ln3 / 2

    Is that right? Something just doesn't feel right.

    EDIT: K, nvm. That first part was right.
    But I didn't get the inverse right.

    x^2=3-e^2y
    x^2 - 3 = -e^2y
    ln(x^2 - 3) = -2y
    -ln(x^2 - 3) / 2 = y
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  2. #2
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    Re: Finding domain of a function.

    Quote Originally Posted by xxStrikeback View Post


    Find the domain of f(x).

    Then find the inverse of the function, and the domain of the inverse.

    I know how to find the domain and inverse, but I just have no idea how to do it in this function.

    For domain, I'm guess 3 - e^2x > or = 0

    -e^2x > or = -3
    e^2x < or = 3
    2x < or = ln3
    x < or = ln3 / 2

    Is that right? Something just doesn't feel right.
    Nothing wrong there.  x < \dfrac{\ln(3)}{2} is the correct domain. The range is f(x) \geq 0 (this will come in handy later)


    For the inverse I use the method of saying that f(x) = y and then solve for x in terms of y.

    y^2 = 3-e^{2x} \Leftrightarrow e^{2x} = 3-y^2

    Can you finish off?

    Once done replace x with f^{-1}(x) and y with x and find the domain in the usual way.


    You can check your answer due to the fact that the domain of f^{-1}(x) is the range of f(x) and vice versa.
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    Re: Finding domain of a function.

    Check the first post, sorry, I edited, I assume, as you were posting.

    x^2=3-e^2y
    x^2 - 3 = -e^2y
    ln(x^2 - 3) = -2y
    -ln(x^2 - 3) / 2 = y

    I submitted that, but apparently it was not right.
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    Re: Finding domain of a function.

    Quote Originally Posted by xxStrikeback View Post
    Check the first post, sorry, I edited, I assume, as you were posting.

    x^2=3-e^2y
    x^2 - 3 = -e^2y
    This bit is good

    Quote Originally Posted by xxStrikeback View Post
    ln(x^2 - 3) = -2y
    -ln(x^2 - 3) / 2 = y

    I submitted that, but apparently it was not right.
    You need to divide by -1 before you take the logarithm since the logarithm is not defined (in the real numbers) for negative arguments and -e^{2y} is always negative so if you try to do what you did you end up trying to take the log of a negative number.

    x^2-3 = -e^{2y}

    -(x^2-3) = e^{2y} \Leftrightarrow 3-x^2 = e^{2y}

    Then continue as normal to get f^{-1}(x) = \dfrac{1}{2}\ln(3-x^2)
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    Re: Finding domain of a function.

    Cheers, got it.

    Now you mention earlier that the range of f(x) was > or = 0. So this means the domain of the inverse should be [0,inf), right?

    When I put this in, it tells me that the [, 0, and ) are right, but the infinity is wrong.

    So I'm confused now on finding the domain of the inverse. What value can x not equal?
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    Re: Finding domain of a function.

    Quote Originally Posted by xxStrikeback View Post
    Cheers, got it.

    Now you mention earlier that the range of f(x) was > or = 0. So this means the domain of the inverse should be [0,inf), right?
    I think I was incorrect earlier when stating that bit, got carried away after the first sentence!

    Quote Originally Posted by xxStrikeback View Post
    When I put this in, it tells me that the [, 0, and ) are right, but the infinity is wrong.

    So I'm confused now on finding the domain of the inverse. What value can x not equal?
    For the domain of the inverse 3-x^2 > 0 so |x| < \sqrt{3} which is the same as -\sqrt{3} < x < \sqrt{3}
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    Re: Finding domain of a function.

    Ah makes sense. Thank you very much.

    While on inverses, I have a question on using f(x) = x^3 + 4x + 2 to find f^-1(2) = ?

    I'm not really sure how to get the inverse here. All I've really done is.

    x=y^3 + 4y +2
    x-2 = y(y^2+4)
    x-2 / y = y^2 + 4

    End up just coming back to the original equation, just not sure how to get it with the x, or rather, y^3.
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    Re: Finding domain of a function.

    Quote Originally Posted by xxStrikeback View Post
    Ah makes sense. Thank you very much.

    While on inverses, I have a question on using f(x) = x^3 + 4x + 2 to find f^-1(2) = ?

    I'm not really sure how to get the inverse here. All I've really done is.

    x=y^3 + 4y +2
    x-2 = y(y^2+4)
    x-2 / y = y^2 + 4

    End up just coming back to the original equation, just not sure how to get it with the x, or rather, y^3.
    The inverse is definitely not pretty so I'm thinking there must be some kind of shortcut involved, possibly using the fact that:

    f(f^{-1}(2)) = 2

    Other than that I'm not sure I'm afraid >.<
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    Re: Finding domain of a function.

    Lol, here's the full problem.

    Properties of Inverse Functions Theorem: Suppose f and g are inverse functions.

    The range of f is the domain of g and the domain of f is the range of g.
    f(a) = b if and only if g(b) = a.
    (a, b) is on the graph of f if and only if (b, a) is on the graph of g.

    The function
    f(x) = x3 + 4x + 2
    is one-to-one. Since finding a formula for its inverse is beyond the scope of this textbook, use Properties of Inverse Functions Theorem to help you compute
    f −1(2).

    Yeah I tried to solve it by hand and then just WolframAlpha'd it and was in shock. I don't know if any of those properties can be used for this?
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