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Math Help - Need help with a few math problems

  1. #1
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    Need help with a few math problems

    I had trouble with these four questions any help would be great!

    1. Identify the center and radius of the circles represented by the following questions

    (a) 3x - √3x+3y-y+ 1/12=0 (b) 12x + 12y+42y=0


    2. Let f(x)= 1/√x-2 and g(x)=√x+1.
    (a) Find the domain of each function.
    (b) Find the domain of (g/f)(x)

    3. Determine whether the functions are even, odd, or neither.

    (a) f(x)= x/√x+1 (b) g(x)=x - √x-7


    4. Let f(x)= x^5/120- x/6 + x.

    (a) Find the local minima and maxima.
    (b) Determine where the function is increasing and decreasing.

    *Sorry I didn't know how to put them in correctly but the / is divided by and the ^5 is to the fifth power and the √ is the square root

    Any help is greatly appreciated
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  2. #2
    Senior Member Peritus's Avatar
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    1. Identify the center and radius of the circles represented by the following questions

    (a) 3x - √3x+3y-y+ 1/12=0
    The idea is to bring the equation to the canonical form of the circle equation by completing the squares:

    <br />
\begin{gathered}<br />
  3x^2  - \sqrt 3 x + 3y^2  - y + \frac{1}<br />
{{12}} = 0 \hfill \\<br />
   \Leftrightarrow x^2  - \frac{1}<br />
{{\sqrt 3 }}x + y^2  - \frac{1}<br />
{3}y + \frac{1}<br />
{{36}} = 0 \hfill \\ <br />
\end{gathered}


    \begin{gathered}<br />
   \Leftrightarrow \left( {x - \frac{1}<br />
{{2\sqrt 3 }}} \right)^2  - \frac{1}<br />
{{12}} + \left( {y - \frac{1}<br />
{6}} \right)^2  - \frac{1}<br />
{{36}} + \frac{1}<br />
{{36}} = 0 \hfill \\<br />
   \Leftrightarrow \left( {x - \frac{1}<br />
{{2\sqrt 3 }}} \right)^2  + \left( {y - \frac{1}<br />
{6}} \right)^2  = \frac{1}<br />
{{12}} \hfill \\ <br />
\end{gathered}
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  3. #3
    Senior Member Peritus's Avatar
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    3. Determine whether the functions are even, odd, or neither.

    (a) f(x)= x/√x+1 (b) g(x)=x - √x-7
    a function f(x) is even if for every x f(x) = f(-x)
    a function f(x) is odd if for every x f(-x) = -f(x)
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  4. #4
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    For the first one I'm confused as to how you find the center and the radius?
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  5. #5
    Senior Member Peritus's Avatar
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    Quote Originally Posted by romanianxromo View Post
    For the first one I'm confused as to how you find the center and the radius?
    There's no reason to be confused that is unless you're not familiar with the canonical form of the circle equation:

    <br />
\left( {x - a} \right)^2  + \left( {y - b} \right)^2  = R^2 <br />

    where (a,b) is the coordinates of the center and R is the radius of the circle.
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