What is the domain of the function y=ln(x^2-3x+2) ?

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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   ...

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neela | Student

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).

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giorgiana1976 | Student

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).

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