Find the domain of 1/sqrt(3x-2) for the expression to produce a real number?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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The domain means that we need to obtain x values in which the functions has real numbers value.

The function 1/sqrt(3x-2) has no defined value when sqrt(3x-2)= 0 and when (3x-2)<0 (negative values)

Now, sqrt (3x-2) = 0

==> 3x-2= 0

==> x= 2/3

then the function is undefined when x= 2/3 and x< 2/3

or the interval (-inf, 1/2] (the function has no real value or does not belong to (R)

so the domain of the function is R- (-inf, 2/3] = (2/3, inf)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we have to notice that the domain for 1/sqrt (3x-2) is the same with the domain for sqrt (3x-2), excepting the values for x which is cancelling the denominator.

Let's find this excepted value for x (we've considered from the beginning that it's just a single value, based on the fact that the expression is a linear equation, with a single solution).

3x-2 = 0

We'll add 2, both sides of the equation:



We'll divide by3, both sides:

x = 2/3.

So, the excluded value for x = 2/3.

Now, let's find the domain for sqrt(3x-2).

For sqrt(3x-2) to exists, the expression (3x-2)>0

So, reiterating the same steps to find the excepted value for x, we'll find that x>2/3.

So the domain of definition is the ineterval (2/3, inf).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the domain of 1/sqrt(3x-2).


Let f(x) = 1/sqrt(3x-2).

For f(x) to be real, the expression under the square root, (3x-2) should be positive.

Therefore 3x-2 > 0. Or 3x > 2. Or x > 2/3. So  x should be non negative . Or

X should belong to the open interval (2/3 , infinity ).

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