# Determine the domain of definition of the function y=ln(x^2-3x+2) .

### 2 Answers | Add Yours

To determine the domain of definition of the function y=ln(x^2-3x+2).

The domain of the of the function y=ln(x^2-3x+2), is the set of all the values of x for which y=ln(x^2-3x+2) is defined.

ln(x^2-3x+2) is defined only when x^2-3x+2 > 0.

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

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

=> x> 2, or x < 1, when the product (x-2)(x-1) > 0.

Therefore the domain of the function set of values of x is (-infinity , 1) or (2 , infinity).

In the domain of definition of the given function has to be all the values of x for the logarithmic function to exist.

We'll impose the constraint 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 = (3-1)/2

x2 = 1

The expression is positive over the intervals:

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

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

**The domain of definition of the given function y=ln(x^2-3x+2) is the reunion of intervals (-infinite , 1) U (2 , +infinite).**