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

Question

Ethan Adkins · Accepted Answer

See below.

A radical's value cannot be negative, or else complex solutions would result. Thus, #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)#.

Abigail Allen · Answer

undefined To find the domain and range of the function $ f(x) = \sqrt{2x^2 - 1} $, we need to consider the restrictions on the input $ x $ that would make the function undefined, as well as the possible output values.

### Domain:
The domain of $ f(x) $ consists of all the possible values that $ x $ can take without making the function undefined. For a square root function, the expression inside the square root must be non-negative.

So, we set $ 2x^2 - 1 \geq 0 $ to find the domain:

$$ 2x^2 - 1 \geq 0 $$
$$ 2x^2 \geq 1 $$
$$ x^2 \geq \frac{1}{2} $$

Taking the square root of both sides, we get:

$$ x \geq \pm \sqrt{\frac{1}{2}} $$
$$ x \geq \pm \frac{1}{\sqrt{2}} $$
$$ x \geq \pm \frac{\sqrt{2}}{2} $$

So, the domain of $ f(x) $ is $ x \geq -\frac{\sqrt{2}}{2} $ and $ x \leq \frac{\sqrt{2}}{2} $.

### Range:
The range of $ f(x) $ consists of all the possible output values of the function. For a square root function, the output is always non-negative (since the square root of a number is non-negative).

So, the range of $ f(x) $ is $ f(x) \geq 0 $.

### Summary:
- **Domain:** $ x $ is in the interval $ x \geq -\frac{\sqrt{2}}{2} $ and $ x \leq \frac{\sqrt{2}}{2} $
- **Range:** $ f(x) $ is in the interval $ f(x) \geq 0 $