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]**

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.**

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