Question

question 1
0.5 pts
which of the following is the domain of the function
$f(x)=\sqrt3{x^{2}-6x + 5}$
check the correct alternative.
$\bigcirc(1,5)$
$\bigcirc(-\infty,1)\cup(5,\infty)$
$\bigcirc(-\infty,1\cup5,\infty)$
$\bigcirc1,5$
$\bigcirc(-\infty,\infty)$

Explanation:

Step1: Recall the domain of cube - root functions

The domain of a function \(y = \sqrt[3]{u}\) is all real numbers because for any real number \(u\), the cube - root \(\sqrt[3]{u}\) is defined. In the function \(f(x)=\sqrt[3]{x^{2}-6x + 5}\), the expression inside the cube - root is \(u=x^{2}-6x + 5\). Since \(y=\sqrt[3]{u}\) has a domain of \((-\infty,\infty)\) for any real - valued \(u\), and \(u=x^{2}-6x + 5\) is a polynomial (and polynomials are defined for all real \(x\)).

Answer:

\((-\infty,\infty)\)