f(x)=1/(4x^2-1)

The domain is x values in which the function is defined.

Since the function is a root, then the denominator (4x^2-1) should not be zero.

4x^2-1=0

==> (2x-1)(2x+1)=0

==> x= 1/2

and x= =1/2

Then the function domain is all R values except for -1/2 and 1/2

The domain = R-{1/2, -1/2}

To determine the domain of f(x) = 1/(4x^2-1)

Solution:

The domain of x is the set of all those values of x for which f(x) is real and defined.

Therefore it is sufficient to examine if the rational expression becomes zero for any x= c in the denominator, where the expession has an unbounded jump and a discontinuity and f(c) becomes undefined.

So the denomonator 4x^2-1 = 0, when 4x^2 = 1, when x^2 = 1/4.

So x = +sqrt(1/4) =1/2 and x = -sqrt(1/4) = -1/2 are the points where f(-1/2) and f(1/2) are undefined.

So x <-1/2 and x>1/2

Before establishing the domain of the function, we notice that the expression of the function is a ratio and we'll set the domain knowing the fact that the division by 0 is not allowed.

For this reason, we'll find out first the x values, for the denominator is cancelling.

4x^2-1 = 0

Being a difference of squares, we'll write as:

4x^2-1 = (2x-1)(2x+1)

(2x-1)(2x+1)=0

We'll put each factor from the product 0:

2x-1=0

2x=1

x=1/2

2x+1=0

2x=-1

x=-1/2

From here we conclude that the function is not defined for x=-1/2 or x=1/2.

So, the domain of definition is:

(-inf,-1/2)U(1/2,+inf)

or

R-{-1/2 , 1/2}