# Find the domain of the given function. Write your answers in both set builder and interval notation, also with absolute value. : f(x)= 2x`/` `sqrt(6x^2 -1)`

aruv | High School Teacher | (Level 2) Valedictorian

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Find the domain of the given function.

`f(x)=(2x)/sqrt(6x^2-1)`

Assume f is real function.

function f is defined if `sqrt(6x^2-1)!=0`

also f is real therefore

`6x^2-1>0`

`6x^2>1`

`x^2>1/6`

`|x|>sqrt(1/6)`

i.e

`-1/sqrt(6)>x and x>1/sqrt(6)`

Thus domain of the function f is

`(-oo,-1/sqrt(6))uu(1/sqrt(6),oo)`

In set builder notation

`D={x: x<-1/sqrt(6) and x>1/sqrt(6),x inRR}`

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

For radical functions of even index defined over real numbers, the domain is the set of possible values. These values must be non-negative.

Again, there is a restriction of division by zero.

Hence `6x^2-1gt0`

`6x^2gt1`

`x^2gt1/6`

`xgtsqrt(1/6)`

So, the domain of `(2x)/(sqrt6x^2-1)` is:

`{x|x in RR, xgtsqrt(1/6)}` orĀ  ` [sqrt(1/6),oo)` .