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

The range of a function is the interval in which the function takes values, or said in other words the set of values that the function can take.

Since a square root function of type

`g(x) =sqrt(x)`

is defined only for positive `x`, it can take only positive values so that it has the range in the interval `[0,+oo)`

Thus the denominator of our function

`h(x) =sqrt(1-4x^2)`

being a square root function takes only positive values for its domain of definition (whichever it should be). Hence, the range of the denominator is `[0,+oo)` .

Because of this, the range of the original function

`f(x) =1/(h(x))`

will be also the interval `[0,+oo)` , or said different the set of all positive real numbers including zero `{x in RR|x>=0}`

(When `h(x) -> +0, f(x) ->+oo` and when `h(x) ->+oo, f(x)->+0` )

**Answer: the range if the function is `[0,+oo)` or equivalent `{x in RR|x>=0}` **

I was just in a hurry to answer such a simple question and did not observed that the function

h(x) = sqrt(1-4x^2)

`h(x) =sqrt(1-4x^2)`

can not take values bigger than 1 because of the square of x,

so that

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

can not take values less than 1.

But the answer below is also mistaking when limiting the

range upper to 3.

What happens when `x¨->+1/2` from left, or when `x-> -1/2` from right? The answer is `sqrt(1-4x^2) ->+0` and `f(x) =1/sqrt(1-4x^2) ->+oo`

**The correct range should be `[1,+oo)` **

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

Solve

`1-4x^2>0`

`4x^2<1`

`x^2<1/4`

`|x|<1/2`

Thus domain of f is

`S={x:|x|<1/2,x in RR}`

Range of f

`1>sqrt(1-4x^2) AA x inR`

So in particular |x|<1. Let us plot graph

Thus range of f

`R={x:1<=x<3 }`