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the domain is x=values in which the function is identified.
to find possible x values we need to solve the function.
so, (x^2-3x+2)^1/2 = 0
==> x = 2, and x=1
Now the function is undefined when (x-2)<0 AND x-1<0
then when x-2 <0 ==> x< 2 which is the interval (-inf,2)
x-1<0 ==> x<1 which is the interval (-inf,1)
Then, the function is undefined when x= (-inf, 2) intersection (-inf,1) = (1,2)
then the functin is defined when (-inf,1] U [2, inf)
To determine the domain of(x^2-3x+2)^(1/2).
Let f(x) = (x^2-3x+2) ^(1/2).
For f(x) to be real the expression iumder the pexponet (1/2) or square root must be positive.
x^2-3x+2 > = 0,
Factorising the left: (x-2)(x-1) > = 0. (x-2)(x-1) is positive only if x<=1 when both factors are negative and so the product is positive. Or x > = 2,
So the domain for x is x<= 1 or x >= 2
The domain of sqrt(x^2-3x+2) is formed from the xalues of x for the expression sqrt(x^2-3x+2) is defined.
To find the domain of sqrt(x^2-3x+2), we must impose the constraint (x^2-3x+2)>0.
To solve the inequality, we'll factorize the quadratic expression:
x^2-3x+2 = x^2 - x - 2x + 2 = 0
We'll factorize the first 2 terms and the next 2 terms, so that:
x(x-1) - 2(x-1) = 0
We'll factorize again and we'll obtain:
We'll set each factor eqaul to 0.
After finding the zeros, we have to test the signs over the 3 intervals:
(-inf,1), (1,2), (2, inf).
We'll choose values from each interval and thest the sign:
For x=0, in interval (-inf,1).
sqrt(0^2-3*0+2) = sqrt2>0
For x=1.5, in interval (1,2)
sqrt(1.5^2-3*1.5+2)=sqrt(2.25 - 4.5 + 2)=sqrt(-0.25) undefined
For x=3, in interval (2, inf)
After testing, we can establish the domain:
Because of the fact that the expression is defined for values of x in the intervals (-inf,1) or (2, inf) and undefined for values of x in the interval (1,2), the domain will be:
Domain: (-inf,1] U [2, inf)
Note: The symbol "U" means reunion of intervals.
We've included also the values x=1 and x=2, because they are the zeros of the expression sqrt(3^2-3*3+2).
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