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Math Help - Domain of Function;Graphing;Zeros algebraically

  1. #1
    Newbie Ai Ekio's Avatar
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    Question Domain of Function;Graphing;Zeros algebraically

    I have four problems I need help with.

    The first two are-
    Find the domain of the function:
    √25-x2
    Edit: After attempting this problem,I've gotten (-5
    ,5) ..is that right?

    and
    Find the domain of the function:
    x/x2-x-6
    Edit: For this one I've gotten (-∞,-2)U(-2,3)U(3,∞).Once again,I'd appreciated some double checking from someone.

    The next is-
    Find the zeros of the function algebraically:
    f(x)=x3-x2-25x+25
    Edit: And for this one,I've gotten 5,-5,1.Correct?

    The last is-
    Graph the function:
    x2-2, x<-2
    f(x)= 5 , -2< x <0

    8x-5, x >0


    ^This one specifically I really need help understanding. ):



    Last edited by Ai Ekio; August 31st 2012 at 01:53 AM.
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  2. #2
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    Re: Domain of Function;Graphing;Zeros algebraically

    For your first check, the √25-x2 is not limited in any domainas it is a parabola opening downward. If graphing is not available simply look at the equation... is there any value of x that would not allow it to be a real number? because x can only be squared, there are no limits.

    Again, with your second problem there are no limits on what x can be (which is what a domain is) except that it cannot be zero because if it was zero, then the fraction would become undefined. That would get you a domain of all reals excluding zero because zero squared would equal zero.

    Your third is correct

    This last one is called a piecewise function, where each segment if the function has the properties of a different toolkit. so, to help you understand, i will write out how that equation would sound. "y will equal X squared minus 2 when X is less than negative 2". that is the first part. So wherever x is less than -2, you would graph x^2-2, and then stop at -2 with an open circle because it cannot actually equal -2. The next would read like "F of X equals 5 (Y=5) when x is greater than or equal to -2 and less than of equal to 0" same deal. Finally, " will equal 8x minus 5 when x is greater than zero (not including zero)" to symbolize an end that includes the final value is a closed circle.

    Hope this has helped
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  3. #3
    Newbie Ai Ekio's Avatar
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    Re: Domain of Function;Graphing;Zeros algebraically

    My understanding was that the restriction for the first problem is anything that results in a negative root,because a negative root isn't possible (without getting into imaginary numbers).And for the second,my reasoning was that when -2 is plugged into the denominator you get 4+2-6,which would be zero and when you plug in 3 you get 9-3-6,which again is a zero;while anything below -2,above 3,and inbetween excluding those numbers would result in either a negative denominator or a positive one,hence my interval notation.
    I believe your last explanation does help me understand better (especially where the closed/open circles are concerned) and I'm trying very hard to understand,so thank you for replying back!
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    Re: Domain of Function;Graphing;Zeros algebraically

    Quote Originally Posted by Ai Ekio View Post
    My understanding was that the restriction for the first problem is anything that results in a negative root,because a negative root isn't possible (without getting into imaginary numbers).And for the second,my reasoning was that when -2 is plugged into the denominator you get 4+2-6,which would be zero and when you plug in 3 you get 9-3-6,which again is a zero;while anything below -2,above 3,and inbetween excluding those numbers would result in either a negative denominator or a positive one,hence my interval notation. I believe your last explanation does help me understand better (especially where the closed/open circles are concerned) and I'm trying very hard to understand,so thank you for replying back!
    I think that FxAperture read the question as \sqrt{25}-x^2 and not as \sqrt{25-x^2}.
    That is because you failed to use any grouping symbols at all in the entire post.
    For example x/x^2-x-6 should be read as \frac{x}{x^2}-x-6, but you meant \frac{x}{x^2-x-6}. So group things.
    Thanks from Ai Ekio
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  5. #5
    Newbie Ai Ekio's Avatar
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    Re: Domain of Function;Graphing;Zeros algebraically

    Quote Originally Posted by Plato View Post
    I think that FxAperture read the question as \sqrt{25}-x^2 and not as \sqrt{25-x^2}.
    That is because you failed to use any grouping symbols at all in the entire post.
    For example x/x^2-x-6 should be read as \frac{x}{x^2}-x-6, but you meant \frac{x}{x^2-x-6}. So group things.
    Oh,okay.Sorry,I'm not very literate when it comes to how to type out equations.Thank you.
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  6. #6
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    Re: Domain of Function;Graphing;Zeros algebraically

    Hello, Ai Ekio!

    FxAperture did an excellent job explaining the piecewise function.
    I'll baby-step through its graphing . . .


    \text{Graph the function: }\:f(x) \;=\;\begin{Bmatrix}x^2-2 && x < -2 \\ 5 && -2 \le x \le 0 \\ 8x-5 && x > 0 \end{array}

    The first function is a parabola: , y \:=\:x^2-2
    Code:
                  |
         *        |        *
                  |
                  |
                  |
          *       |       *
                  |
                  |
           *      |      *
                  |
            *     |     *
        -----*----+--------
               *  |  *
                  *
                  |
    But we want only the portion less than -2.
    Code:
                  |
         *        |          .
                  |
                  |
                  |
          *       |       .
                  |
                  |
           o      |      .
           :      |
           :.     |     .
        ---+-.----+----.---
          -2   .  |  .
                  .
                  |
                  
                  |
                 o|
                  |

    The second function is: . y \:=\:5, a horizontal line.
    Code:
                  |
                  |
          * * * * * * * * * 
                  |
                  |
                  |
                  |
        ----------+--------
                  |
                  |
    But we want the portion between -2 and 0, inclusive.
    Code:
                  |
                  |
           ********
           :      |
           :      |
           :      |
           :      |
        ---+------+--------
          -2      |
                  |

    The third function is a line.
    Code:
                  |
                  |      *
                  |
                  |     *
                  |
                  |    *
                  |
                  |   *
                  |
                  |  *
                  |
        ----------+-*------
                  |
                  |*
                  |
                  *
                  |
                 *|
                  |
    But we want the portion greater than zero.
    Code:
                  |
                  |      *
                  |
                  |     *
                  |
                  |    *
                  |
                  |   *
                  |
                  |  *
                  |
        ----------+-*------
                  |
                  |*
                  |
                  o
                  |
                  |
                  |


    Therefore, the graph looks like this:

    Code:
                  |
         *        |      *
                  |
                  |     *
                  |
          *       |    *
           ********
                  |   *
           o      |
           :      |  *
           :      |
        ---+------+-*------
          -2      |
                   *
                  |
                  o
                  |
    Thanks from Ai Ekio
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  7. #7
    Newbie Ai Ekio's Avatar
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    Re: Domain of Function;Graphing;Zeros algebraically

    Quote Originally Posted by Soroban View Post
    Hello, Ai Ekio!
    Thank you!The visuals really helped. (:
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