# What is the range of f(x) = 1 + sqrt(9 - x^2)?

May 10, 2017

$1 \le f \left(x\right) \le 4$

#### Explanation:

The values that $f \left(x\right)$ can take are dependent on the values for which $x$ is defined.

So, in order to find the range of $f \left(x\right)$, we need to find its domain and take evaluate $f$ at these points.

$\sqrt{9 - {x}^{2}}$ is only defined for $| x | \le 3$. But since we're taking the square of $x$, the smallest value it can take is $0$ and the largest $3$.

$f \left(0\right) = 4$

$f \left(3\right) = 1$

Thus $f \left(x\right)$ is defined over $\left[1 , 4\right]$.