What is the domain of the function y=ln(x^2-3x+2) ?
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calendarEducator since 2008
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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 ...
(The entire section contains 2 answers and 254 words.)
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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).
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