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 $[0,oo)$ .

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