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We need to find the domain of the function f(x) = sqrt(x-2)/sqrt(3-x).
The domain of a function is all values of x for which the function has a real value.
For f(x) to be real the numerator and the denominator have to be real. And the denominator cannot be equal to 0 as that would make the value of f(x) undefined.
This means x - 2 cannot be less than 0 and 3-x has to be greater than 0.
So we have x - 2 >=0 and 3-x > 0
=> x >= 2 and 3 > x
=> x >= 2 and x < 3
Therefore the domain of the function is all values of x that lie in [ 2, 3).
Given the function:
f(x) = sqrt(x-2) / sqrt(3-x) we need to find the domain.
Since the function if a quotient, then the denominator can not be zero.
Also, the square root must be a values greater than zero.
Then the domain is all x values such that the numerator is equal or greater than zero and the denominator is greater than zero.
Let us find the domain of the denominator:
==> sqrt(3-x) > 0
==> 3-x > 0
==> -x > -3
==> x < 3
==> x belongs to the interval ( -inf, 3)..........(1)
Now we will determine the domain for the numerator.
==> sqrt(x-2) >= 0
==> x-2 >= 0
==> x >= 2
==> x belongs to the interval [2, inf)............(1)
Then we conclude that the domain of f(x) is (1) and (2).
==> x belongs to the interval (-inf, 3) and [2, inf) = [2,3)
==> x = [ 2, 3 )
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