# How do you find the domain and range of y =sqrt(x-2) / (6+x)?

Apr 24, 2015

You need to ensure that:
1] the denominator is different from zero, so:
$6 + x \ne 0$
$x \ne - 6$
2] the argument of the square root must be bigger or equal to zero, so:
$x - 2 \ge 0$
$x \ge 2$
These conditions are used to avoid points or intervals where your function cannot be calculated.
Finally the domain is all real $x$ bigger or equal to $2$.

For the range I evaluated the maximum through the derivative:
$y ' = \frac{10 - x}{2 \cdot \left(\sqrt{x - 2}\right) \cdot {\left(x + 6\right)}^{2}}$
The maximum is at $x = 10$ (by setting $y ' = 0$) corresponding to $y = \frac{\sqrt{8}}{16}$ giving a range for $y$ from zero to $\frac{\sqrt{8}}{16}$.