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Math Help - Trigonometric and domain/range question.

  1. #1
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    Trigonometric and domain/range question.

    Hello everyone! I just need help with two of these questions on my math homework.

    The first question is:

    For b > 0, what are the domain and range of b^x?

    The second question is:



    Please explain it as well because I would actually like to understand how the problems were solved as opposed to just being given an answer.
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  2. #2
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    Quote Originally Posted by FullBox View Post

    For b > 0, what are the domain and range of b^x?
    For domain, what values can x take?

    For range, what values will the function output? Are there any asymptotes here?
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  3. #3
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    For your second question.. Use SOHCAHTOA to find the lengths of a,h and k.
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    Quote Originally Posted by pickslides View Post
    For domain, what values can x take?

    For range, what values will the function output? Are there any asymptotes here?
    I am so sorry, I forgot to type a part of the problem. Here is what I missed.

    For b > 0, what are the domain and range of f(x) = b^x?

    Quote Originally Posted by guyonfire89 View Post
    For your second question.. Use SOHCAHTOA to find the lengths of a,h and k.
    I am not sure what that abbreviation stands for.
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  5. #5
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    Quote Originally Posted by FullBox View Post
    Here is what I missed.

    For b > 0, what are the domain and range of f(x) = b^x?
    The additional information does not change my response.


    Quote Originally Posted by FullBox View Post

    I am not sure what that abbreviation stands for.
    SOH means sine theta = opposite/hypotenuse

    CAH means cosine theta = adjacent/hypotenuse

    TOA means tangent theta = opposite/adjacent
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  6. #6
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    Quote Originally Posted by pickslides View Post
    The additional information does not change my response.
    Thats exactly how the problem was phrased, though. There was no additional information included. I'll even snapshot it so that you can make sure.



    Oh, and for the second problem, I have to express my answer in terms of a, h, and k, not solve for them. Thats what confused me; the fact no numbers were involved [in expressing the sides].

    For example, on the left triangle, sin75 = h/a, cos15 = h/a.
    On the right triangle, sin45 = k/a and cos45 = a/k. That would make the tangent = (k/a) / (a/k), I'm not sure if I can simplify it further (maybe [k/a]^2). That gives me the base for the right triangle. I'm still totally lost on the left triangle though.

    However, what would I do with that information? Since its expressed in letters, how would I solve for the side that does not have a letter assigned to it?
    Last edited by FullBox; September 16th 2010 at 03:38 AM.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    1.

    Try using x = -infinity, x = 0 and x = infinity. Those three points often help in determining the domain.


    By the way, do you know the shape of the graph  y = e^x ? This might help you.

    2. For this one, I think there is a simpler way, that is using Pythagoras' Theorem. Do you know about it?
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    1.

    Try using x = -infinity, x = 0 and x = infinity. Those three points often help in determining the domain.


    By the way, do you know the shape of the graph  y = e^x ? This might help you.
    I just googled the graph to make sure, and yes I've seen it before. Based on it, it looks like the range is  y > 0 since it never falls below zero on the left side and shoots up infinitely on the right.

    The domain seems to be all real numbers, or negative infinity to positive infinity.

    2. For this one, I think there is a simpler way, that is using Pythagoras' Theorem. Do you know about it?
    Yeah, I know the pythagorean theorem. I thought of that at first but I don't know how I would express the answer if one of the sides on both triangles is unidentified.

     a^2 = h^2 + x^2
     a^2 = k^2 + x^2

    In both cases though, x is not identified/given a letter, so I don't know how to express the answer in terms of only h, a, and k.

    Would I do (a^2)/(h^2) and (a^2)/(k^2) ?

    Would the answer result in  (2a^2) / ((k^2) + (h^2))
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  9. #9
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    1. Yes, the domain is negative infinity to infinity.

    The range, however is y > 0. The graph never touches the x-axis.

    2. Be careful!

    Let y be the longer length, and x be the shorter length.

    Then;

    a^2 = h^2 + x^2

    a^2 = k^2 + y^2

    x + y is what you're looking for.

    So,

    x = \sqrt{a^2 - h^2}

    y = \sqrt{a^2 - k^2}

    x + y = \sqrt{a^2 - h^2} + \sqrt{a^2 - k^2}

    That's not too easy on the eyes though. Let's take trigonometry.

    cos(75) = \frac{y}{a}

    cos(45) = \frac{x}{a}

    So;

    y = acos(75)

    x = acos(45)

    x + y = a(cos(75) + cos(45))
    Last edited by Unknown008; September 16th 2010 at 04:20 AM. Reason: math tag problem in line of the two pythagorean euqations
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  10. #10
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    Quote Originally Posted by Unknown008 View Post
    1. Yes, the domain is negative infinity to infinity.

    The range, however is y > 0. The graph never touches the x-axis.

    2. Be careful!

    Let y be the longer length, and x be the shorter length.

    Then;

    a^2 = h^2 + x^2\\<br /> <br />
a^2 = k^2 + y^2

    x + y is what you're looking for.

    So,

    x = \sqrt{a^2 - h^2}

    y = \sqrt{a^2 - k^2}

    x + y = \sqrt{a^2 - h^2} + \sqrt{a^2 - k^2}

    That's not too easy on the eyes though. Let's take trigonometry.

    cos(75) = \frac{y}{a}

    cos(45) = \frac{x}{a}

    So;

    y = acos(75)

    x = acos(45)

    x + y = a(cos(75) + cos(45))
    Thank you very much! So what you did was use the pythagorean theorem to get the answer, but then simplified it into trigonometric terms. Showing the steps you did really helped me understand how you got the answer, thanks a ton man!
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  11. #11
    MHF Contributor Unknown008's Avatar
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    I got the width of the alley in terms of a, h and k first using Pythagoras' Theorem.

    Then, I got the width of the alley by using trigonometry, but then, it's only in terms of a.

    Those are two different methods.

    I could have made use of other trigonometric ratios.

    For example tan.

    tan(45) = \dfrac{k}{y}

    tan(75) = \dfrac{h}{x}

    Alley width = x + y = \dfrac{h}{tan(75)} + \dfrac{k}{tan(45)}

    With sin.

    sin(45) = \dfrac{y}{a}

    sin(15) = \dfrac{x}{a}

    Alley width = x + y = a(sin(45) + sin(15))
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