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

Dec 2, 2017

Range: $y \ge 0$
Domain: $x \ge 1$ and $x \le - 1$

#### Explanation:

We can first consider the range, rather simply, we must consider all the values that $y$ can take on, but as we know $\sqrt{\alpha} \ge 0$ ,$\alpha \in \mathbb{R}$, or in other words, the square root of a value can never be negative for $x \in \mathbb{R}$, so hence $y \ge 0$

Now we can consider the domain, to what we can consider for what values of $x$ yields and valid value of $y$, we know the valid values of $\sqrt{\delta}$ is where $\delta \ge 0$, so in this circumstance we must consider where ${x}^{2} - 1 \ge 0$ we can sketch to find the values:
graph{x^2-1 [-3.538, 3.572, -1.437, 2.118]}

we can evidently see that where ${x}^{2} - 1 \ge 0$ is for $x \ge 1$ and $x \le - 1$

So hence the domain is; $x \ge 1 \mathmr{and} x \le - 1$