(x^2-3x+2)^1/2

the domain is x=values in which the function is identified.

to find possible x values we need to solve the function.

so, (x^2-3x+2)^1/2 = 0

==> [(x-2)^1/2][(x-1)^1/2]=0

==> x = 2, and x=1

Now the function is undefined when (x-2)<0 **AND** x-1<0

then when x-2 <0 ==> x< 2 which is the interval (-inf,2)

x-1<0 ==> x<1 which is the interval (-inf,1)

Then, the function is undefined when x= (-inf, 2) intersection (-inf,1) = (1,2)

then the functin is defined when (-inf,1] U [2, inf)

To determine the domain of(x^2-3x+2)^(1/2).

Solution:

Let f(x) = (x^2-3x+2) ^(1/2).

For f(x) to be real the expression iumder the pexponet (1/2) or square root must be positive.

So

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

Factorising the left: (x-2)(x-1) > = 0. (x-2)(x-1) is positive only if x<=1 when both factors are negative and so the product is positive. Or x > = 2,

So the domain for x is x<= 1 or x >= 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).**