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To find the domain of f(x) = log2(x^2-5x+6).
We know the domain of x is the set of values of x for which log2 (x^2-5x+6) is real.
This is when x^2-5x+6 > 0.
Therefore x^2-5x+ (5/2)^2 -(5/2)^2+6 > 0
Therefore (x-5/2)^2 - 6.25+6 > 0.
(x-5/2)^2 > 0.25.
(x-5/2) > (1/2) , Or x-5/2 < -1/2.
Or x > 5/2+1/2 = 3, Or x < 5/2-1/2 = 2.
So If x > 3 , Or x < 2, then x^2-5x+6 > 0, then f(x) is real.
Therefore the domain of the function f(x) is (-infinity , 2) or (3, infinity).
The domain of definition of the given function contains the admissible 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 - 5x + 6 > 0
We'll compute the roots of the expression:
x^2 - 5x + 6 = 0
We'll apply the quadratic formula:
x1 = [5 +/- sqrt(25 - 24)]/2
x1 = (5+1)/2
x1 = 3
x2 = 2
The expression is positive over the intervals:
(-infinite , 2) U (3 , +infinite)
So, the logarithmic function is defined for values of x that belong to the intervals (-infinite , 2) U (3 , +infinite).
The reunion of intervals represents the domain of definition of the given function f(x) = log 2 (x^2 - 5x + 6).
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