# How do you find the domain and range of f(x) = 1 + sqrt(9 - x^2)?

Jun 5, 2017

The domain of $f \left(x\right)$ is all the values for which

$9 - {x}^{2} \ge 0$

$\left(3 - x\right) \cdot \left(3 + x\right) \ge 0$

Which is valid for $x$ in $\left[- 3 , 3\right]$

Hence the domain is ${D}_{f} = \left[- 3 , 3\right]$

For the range of the function we have that

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

and the maximum value of f(x) is achieved when

$9 - {x}^{2}$ is maximized which happens for $x = 0$

and that is $f \left(0\right) = 4$

Hence the range of the function is

${R}_{f} = \left[1 , 4\right]$

The graph of the function is