Question

determine where the function f(x) is continuous.
f(x) = \sqrt3{2 - x}
the function is continuous on the interval .
(simplify your answer. type your answer in interval notation.)

Explanation:

Step1: Recall continuity of cube - root functions

The cube - root function $y = \sqrt[3]{u}$ is continuous for all real values of $u$.
Let $u = 2 - x$. Since $u$ can be any real number for the cube - root function $\sqrt[3]{u}$, there are no restrictions on the value of $x$ for the function $f(x)=\sqrt[3]{2 - x}$.

Step2: Write the interval

The domain of $f(x)$ is all real numbers. In interval notation, the set of all real numbers is $(-\infty,\infty)$.

Answer:

$(-\infty,\infty)$