How do you find the domain and range of #f(x) =sqrt(2x^2-1)#?

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Answer 1 · 2017-08-20T00:41:43

See below.

The value under a radical cannot be negative, or else the solutions will be complex.

So,

#2x^2-1\geq0#

#2x^2geq1#

#x^2geq1/2#

#xgeq\sqrt(2)/2# and #xleq-\sqrt(2)/2#. In interval notation, this is #(-oo,-\sqrt(2)/2] uu[\sqrt(2)/2,oo)#

Since the radical value will always be greater than or equal to zero, the range is just #[0,oo)#.