Which is the domain of (x^2-3x+2)^1/2?

The domain of sqrt(x^2-3x+2) is formed from the xalues of x for the expression sqrt(x^2-3x+2) is defined.

To find the domain of sqrt(x^2-3x+2), we must impose the constraint (x^2-3x+2)>0.

To solve the inequality, we'll factorize the quadratic expression:

x^2-3x+2 = x^2 - x - 2x + 2 = 0

We'll factorize the first 2 terms and the next 2 terms, so that:

x(x-1) - 2(x-1) = 0

We'll factorize again and we'll obtain:

(x-1)(x-2)=0

We'll set each factor eqaul to 0.

x-1=0

x=1

x-2=0

x=2

After finding the zeros, we have to test the signs over the 3 intervals:

(-inf,1), (1,2), (2, inf).

We'll choose values from each interval and thest the sign:

For x=0, in interval (-inf,1).

sqrt(0^2-3*0+2) = sqrt2>0

For x=1.5, in interval (1,2)

sqrt(1.5^2-3*1.5+2)=sqrt(2.25 - 4.5 + 2)=sqrt(-0.25) undefined

For x=3, in interval (2, inf)

sqrt(3^2-3*3+2)= sqrt2>0

After testing, we can establish the domain:

Because of the fact that the expression is defined for values of x in the intervals (-inf,1) or (2, inf) and undefined for values of x in the interval (1,2), the domain will be:

Domain: (-inf,1] U [2, inf)

Note: The symbol "U" means reunion of intervals.

We've included also the values x=1 and x=2, because they are the zeros of the expression sqrt(3^2-3*3+2).