# How do you find the domain and range of f(x)= -2 + sqrt(16 - x^2)?

Nov 24, 2017

Find the highest and lowest x-values for which the function is defined, then find the highest y-value (which will occur at $x = 0$ because that's where $16 - {x}^{2}$ is at its largest), and the lowest y-value (occurs where $16 - {x}^{2} = 0$.

Domain: $- 4 \le x \le 4$
Range: $- 2 \le f \left(x\right) \le 2$

#### Explanation:

Look for where the function is defined. Recall that we cannot take the square root of a negative number; thus, for all x, $16 - {x}^{2}$ must be positive. This will clearly only occur for ${x}^{2} < 16$, aka $\left\mid x \right\mid < 4$

Thus, this function is only defined for $- 4 \le x \le 4$. That is our domain.

The highest $f \left(x\right)$ value will occur when $\sqrt{16 - {x}^{2}}$ is at its highest, which will occur at $x = 0$. At that point, $f \left(0\right) = - 2 + \sqrt{16} = - 2 + 4 = 2$

The lowest $f \left(x\right)$ value occurs when $\sqrt{16 - {x}^{2}} = 0$. At this point we have $f \left(4\right) = - 2 + \sqrt{16 - 16} = - 2$

Thus, our range will be $- 2 \le f \left(x\right) \le 2$