# How do you find the domain and range of sqrt(x-3) - sqrt(x+3)?

Jun 6, 2017

${D}_{f} = x \ge 3$

${R}_{f} = - \sqrt{6} \le f \left(x\right) < 0$

#### Explanation:

In order to find the domain and range, we need to find the values of $x$ for which the function is defined.

Since we have square roots, we know that they both need to be greater than or equal to $0$. Since the first square root is a subtraction, it is more restrictive so it will control the domain.

$\sqrt{x - 3} \ge 0$

$x - 3 \ge 0$

$x \ge 3$

$\therefore {D}_{f} = x \ge 3$

Now the range is a bit more tricky to find, but if we look at out function, we see that there's a subtraction happening, and this is key to finding the range of $x$.

If we look at what is being subtracted, it's $\sqrt{x - 3} - \sqrt{x + 3}$. By looking, we know that this subtraction will always be negative because $\sqrt{x + 3} > \sqrt{x - 3}$. Taking the smallest valid value of $x$ gives us $\sqrt{0} - \sqrt{6} = - \sqrt{6}$. Thus, our function will always be greater than or equal to $- \sqrt{6}$.

Now, for small values of $x$, the constants will be significant in deciding the value of $f$, so $f$ will be more negative as there will be a larger difference between the two square roots. However, as $x$ becomes larger and larger, the constants become less significant and the difference between the square roots becomes smaller. Thus $f$ becomes less negative and approaches $0$.

However, we know that $\sqrt{x - 3} \ne \sqrt{x + 3}$, so the function will never touch $0$. Thus $0$ is the limiting value of $f$.

${R}_{f} = - \sqrt{6} \le f \left(x\right) < 0$