We have to find the domain of the function y= ln (x^2 - 3x + 2).

Now domain refers to all the values of x for which y exists.

Here y will exist for all values of x that allow x^2 - 3x + 2 to be positive as the logarithm for 0 or negative values is not defined.

Therefore x^2 - 3x + 2 >0

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

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

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

Now this is true if (x - 1)>0 and (x - 2)>0.

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

Therefore x > 2

(x - 1) (x - 2) >0 also holds if both (x - 1)< 0 and (x - 2)<0

=> x < 1 and x < 2

Therefore x< 1

**So the domain of the function is all values of greater than 2 and less than 1.**

y= ln (x^2 - 3x + 2 )

We need to find the domain of the fuction y.

The domain of y is all x values where the function is defined.

We know that ln (x^2 - 3x + 2 ) only defined if ( x^2 - 3x + 2 ) is positive.

==> ( x^2 - 3x + 2 ) > 0

Let us factor.

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

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

==> x > 2 AND x > 1

==> x > 2

==> x = ( 2, inf).

OR:

( x-2) < 0 and ( x-1) < 0

==> x < 2 and x < 1

==> x < 1

==> x = ( -inf , 1)

**Then, the domain of the function y is:**

**x= (-inf , 1) U ( 2, inf) **

**OR:**

**x = R - [ 1,2]**

y = ln (x^2-3x+2).

The domain of rhe function is the set of values of x for which the y is defined.

Since the y is logarithm of (x^2-3x+2), y is defined only for (x^2-3x+2)> 0.

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

We know f(x) > 0 , if x is outside the interval (x1,x2) , where x1 and x2 are the roots of x^2-3x+2 = 0.

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

So x1 = 1 and x2 = 2.

So f(x) = (x-1)(x-2) > 0 if <1 and x>2.

Therefore , the domain x of y = ln(x^22-3x+2) is x in (-infinity , 1) or (2 ,ifinity).

In the domain of definition of the given function there are found all the admissible values of x for the logarithmic function to exist.

We'll impose the conditions for the logarithmic function to exist: the argument of logarithmic function has to be positive.

x^2 - 3x + 2 > 0

We'll compute the roots of the expression:

x^2 - 3x + 2 = 0

We'll apply the quadratic formula:

x1 = [3 +sqrt(9 - 8)]/2

x1 = (3+1)/2

x1 = 2

x2 = 1

The expression is positive over the intervals:

(-infinite , 1) U (2 , +infinite)

**So, the logarithmic function exists for values of x that belong to the ranges (-infinite , 1) U (2 , +infinite).**

The reunion of intervals represents the domain of definition of the given function f(x) = ln (x^2 - 3x + 2).