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).