Question

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

Answer 1

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

Answer 2

x^2 - 3x + 2 = 0

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

Do I factor to solve?

Answer 7

\[x^2-3x+2=(x-1)(x-2)\]

Answer 8

has to be greater than or equal to 2

Answer 9

So the domain would be [2, infinity) ?

Answer 10

I think less than 1 works as well

Answer 11

[-inf,1][2,inf]

Answer 12

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

Answer 13

exactly remidia

Answer 14

You're correct, remidia.

Answer 15

thank you!

Answer 16

I guess that implies that I was incorrect

Answer 17

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

Answer 18

Oh, I made a slight mistake there

Answer 19

Can anyone help me with this? I need to find the domain of

\[h(x) = 1/( sinx - (1/2))\]