Question

Question #e484b

Answer 1

`Df=(-oo,-sqrt2/2]uu[sqrt2/2,+oo)`

Explanation:

You didn't respond but i guess this is `sqrt(2x^2-1)` , makes more sense.

`f(x)=sqrt(2x^2-1)`
#f(x)=0 <=> sqrt(2x^2-1)=0 <=> 2x^2-1=0 <=> 2x^2 = 1 <=> x^2=1/2 <=> x=1/sqrt2 or x=sqrt2/2# or `x=-sqrt2/2`

`sqrt2/2 ~= 0.7`

To define the function we need `2x^2-1>=0` `x>=sqrt2/2` or `x<=-sqrt2/2`

so `Df=(-oo,-sqrt2/2]uu[sqrt2/2,+oo)`

More detailed solution