Find the domain: h(x) = 1/ sqroot(x^2-5x) So I got x(x-5) 0 And I understand that x5 ... but I don't get why x0? Thanks.

Question

Find the domain: 

h(x) = 1/ sqroot(x^2-5x)

So I got x(x-5)> 0

And I understand that x>5 ... but I don't get why x<0.. shouldn't it be x>0?

Thanks.

anonymous · Answer

think of what $$y=x^2-5x$$ looks like. 
it is a parabola that opens up with zeros at $x=0$ and at $x=5$ draw a mental picture and you cam see why it is non-negative |dw:1338231131428:dw| on $x\leq 0$ and $x\geq5$

anonymous · Answer

course since it is in the denominator  it cannot be 0 either, so actual answer should be 
$$x<0$$ or 
$$x>5$$

anonymous · Answer

Ohh ok, that makes sense. And just to confirm.. if it was x^2-5x+2 It would be a polynomial and then it's domain would be all real numbers?

anonymous · Answer

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