Domain of a Function with Sign Graph

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Homework Statement



Find the domain of the function. You will need a sign graph. Answer in interval notation.

Homework Equations



f(x)= (3)/[(x2-4x)1/2]

The Attempt at a Solution


No clue - I don't know how to use a sign graph or what the positive/negative results mean as solutions on a number line.
 
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I am not sure what a sign graph is.

But the domain is the x values for f being defined. There is a square root, so what is inside can't be less than zero. Since it is a fraction, the denominator can't be zero.
 
Sine graph?
 
I meant sine graph instead of sign graph. So what do I do? Should I set the denominator to greater than or equal to zero and solve?

I tried solving using the above method but I think my math is wrong. Nonetheless it is wrong because my teacher wants a sine graph on a number line...

1) (sqrt(x2-4x) greater than or equal 0
2) x2-4x greater than or equal 0 (this step is wrong I'm pretty sure - squaring both sides)
3) x2-4x+4 greater than equal 4
4) (x-2)2 greater than equal 4

... ? I still need the sine graph.
 
I suspect "sign graph" was the correct original intent.

When solving x^2-4x = x(x-4) > 0, this boils down to the cases:

\left\{ \begin{array}{r}{ x > 0 \\ x-4<0} \end{array}

\left\{ \begin{array}{rl} {x<0 \\ x-4 >0} \end{array}

This can also be illustrated with a sign graph of x^2 - 4x:

+ + + + + + + + + + + - - - - - - + + + + + + + + +
___________________0________4________________

And from this one can determine the domain...

Solving (x-2)^2 > 4 \Rightarrow |x-2| > 2 is also valid.

--Elucidus

EDIT: Fixed a TeX gaffe.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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