How do I find the domain of
f(x) = (sqrt)(x^2-3x+2) ?

Question

How do I find the domain of 
f(x) = (sqrt)(x^2-3x+2)  ?

across · Answer

A function with a radical in it is not defined where the radicand is less than zero.

hero · Answer

x^2 - 3x + 2 = 0

anonymous · Answer

The equation is : $$f(x) = \sqrt{x^2-3x+2}$$

across · Answer

Instead of solving$$f(x)=\sqrt{x^2-3x+2}$$you solve$$x^2-3x+2\geq0$$That will give you the domain of your function.

hero · Answer

x^2 - 2x - 1x + 2 = 0

x(x-2)-1(x-2) = 0

(x-2)(x-1) =0

x = 2,1

x has to be greater than 2

anonymous · Answer

Do I factor to solve?

across · Answer

$$x^2-3x+2=(x-1)(x-2)$$

hero · Answer

has to be greater than or equal to 2

anonymous · Answer

So the domain would be [2, infinity) ?

anonymous · Answer

I think less than 1 works as well

hero · Answer

[-inf,1][2,inf]

anonymous · Answer

(- inf, 1] U [2, inf) ?

anonymous · Answer

exactly remidia

across · Answer

You're correct, remidia.

anonymous · Answer

thank you!

hero · Answer

I guess that implies that I was incorrect

anonymous · Answer

It does :.( You can't have [-inf, 1][2, inf] as you can never reach infinity

hero · Answer

Oh, I made a slight mistake there

anonymous · Answer

Can anyone help me with this? I need to find the domain of 

$$h(x) = 1/( sinx - (1/2))$$