Question #e484b

1 Answer
Nov 20, 2017

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 solutionenter image source here