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Math Help - Domain Range and Function Graphic

  1. #1
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    Unhappy Domain Range and Function Graphic

    I have been trying all day and I haven't been able to figure this homework problem out. Math is my weakest area. Can someone help me figure it out and maybe tell me how to do it? I would really appreciate it

    1. Sketch a graph of the function: f(x) = 3x-2/√x^3+8

    x cubed plus 8 is all under the radical

    Find the function's domain and range


    I really have big problems with finding domain and range. Will someone tell me what the domain and range is and how they got their answer?? Also, someone told me to use a graphic calculator to graph this but he told us not to and I don't know how to do it anyway. AHH. Help. Homework is due tomorrow and this is the only problem I have left. I really need a good grade or my school won't let me continue this program.
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  2. #2
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    Quote Originally Posted by PerdidoEnElSiglo View Post
    1. Sketch a graph of the function: f(x) = 3x-2/√x^3+8

    x cubed plus 8 is all under the radical

    Find the function's domain and range
    The denominator cannot be zero and the term inside the square root must be positive.

    x^3+8>0
    x^3>-8
    x>-2
    That is the domain.
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  3. #3
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    what about the range? and how do you graph that?
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  4. #4
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    Here is the graph.
    Let us assume the max point is y=1. Then the range is y\leq 1. The problem is that that point is not 1, and I do not know how you are asked to find this. If this is a Calculus problem I can show you but it seems to not and hence it would be useless for me to explain.
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  5. #5
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    Okay well I graphed it because we are allowed to graph it...I just read the instructions for a different problem. I just don't know how to get the range
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  6. #6
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    Quote Originally Posted by PerdidoEnElSiglo View Post
    Okay well I graphed it because we are allowed to graph it...I just read the instructions for a different problem. I just don't know how to get the range
    [-2, infinity)
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    Quote Originally Posted by AfterShock View Post
    [-2, infinity)
    I think you got the wrong graph.
    ---
    The way to get the range is to find the maximum point for that curve, but I do not think the person who is asking knows Calculus. I have another way but it realies on the properties of a cubic, again I do not think he is supposed to know that either.
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    Quote Originally Posted by PerdidoEnElSiglo View Post
    what about the range? and how do you graph that?
    The most direct way is to simply graph the thing and take a look. If you are more brave you can do something more analytical. However, I agree with ThePerfectHacker, if you don't have the Calculus background estimation is your only way. I get the range to be - infinity to about 1.19 or so.

    -Dan
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    Quote Originally Posted by topsquark View Post
    The most direct way is to simply graph the thing and take a look. If you are more brave you can do something more analytical. However, I agree with ThePerfectHacker, if you don't have the Calculus background estimation is your only way. I get the range to be - infinity to about 1.19 or so.

    -Dan
    There is another way how to do it. I can post it.
    If you ever saw how I do limit problems.

    But this one leads to a cubic, and you need to find the values of y that will make it lead to solutions above x=-2.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    I think you got the wrong graph.
    ---
    The way to get the range is to find the maximum point for that curve, but I do not think the person who is asking knows Calculus. I have another way but it realies on the properties of a cubic, again I do not think he is supposed to know that either.
    I apologize. I had Maple graph the wrong thing.

    It's about (-infinity, 1.1832)
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  11. #11
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    Quote Originally Posted by PerdidoEnElSiglo View Post
    I have been trying all day and I haven't been able to figure this homework problem out. Math is my weakest area. Can someone help me figure it out and maybe tell me how to do it? I would really appreciate it

    1. Sketch a graph of the function: f(x) = 3x-2/√x^3+8

    x cubed plus 8 is all under the radical

    Find the function's domain and range


    I really have big problems with finding domain and range. Will someone tell me what the domain and range is and how they got their answer?? Also, someone told me to use a graphic calculator to graph this but he told us not to and I don't know how to do it anyway. AHH. Help. Homework is due tomorrow and this is the only problem I have left. I really need a good grade or my school won't let me continue this program.
    Lets assume you mean:

    f(x) = (3x-2)/\sqrt{x^3+8}

    As x \to +\infty this clearly goes to 0, and it is also well
    behaved down to the bottom end of its domain where it goes to -infinity.

    Therefore is has a calculus type maximum somewere in (-2,\infty)

    So we differentiate f(x) and set the derivative equal to zero and solve to find the x corresponding to the maxima.

    f'(x)=\frac{3}{(x^3+8)^{1/2}} + \frac{(3x-2)(-1/2)(2x^2}{(x^3+8)^{3/2}}

    Setting this equal to zero and simplifying gives:

    x^3 - 2x^2 - 16=0

    Sketching this we see that it has one real root near x=3.4.

    Formula itteration of:

    x_{n+1}=(2 x_{n}^2+16)^{1/3}

    refines the root to x \approx 3.39124 , so the maximum of f(x) is \approx f( 3.39124)\approx1.19225 . (If we wish the exact root we would use the cubic formula to find it in closed from)

    Hence the range is \approx (-\infty,1.19225].
    (note if we had the exact maxima the function achives its maxima that is why I have used a ] closing bracket)

    RonL
    Last edited by CaptainBlack; January 12th 2007 at 12:16 PM.
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    Here is the way I do range problems....

    We need to find all y such that,
    y=\frac{3x-2}{\sqrt{x^3+8}}
    Has a solution in the domain, that is, x>-2.

    y\sqrt{x^3+8}=(3x-2)
    y^2(x^3+8)=(3x-2)^2
    x^3y^2+8y^2=9x^2-12x+4
    x^3y^2-9x^2+12x+(8y^2-4)=0
    Thus, we need to find all y such that this cubic equation has a solution for x>-2.

    This is mine 41th Post!!!
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