Question

what is the range of $f(x) = \sqrt3{x - 4}$? (1 point) \
$\\{x|-\infty \leq x \leq \infty\\}$ \
$\\{y|y \in \mathbb{r}\\}$ \
$\\{x|-\infty < x < \infty\\}$ \
$\\{y|0 < y < \infty\\}$

Explanation:

Step1: Recall the cube root function

The cube root function \( \sqrt[3]{u} \) has a domain and range of all real numbers because we can take the cube root of any real number (positive, negative, or zero), and the result can also be any real number.

Step2: Analyze \( f(x)=\sqrt[3]{x - 4} \)

Let \( u=x - 4 \). The function \( f(x)=\sqrt[3]{u} \) where \( u=x - 4 \). The variable \( u=x - 4 \) can take on any real - number value (since \( x \) can be any real number, \( x-4 \) can be any real number). And since the cube root of any real number \( u \) is also a real number, the range of \( y = f(x)=\sqrt[3]{x - 4} \) is all real numbers. In set - builder notation, the range of a function \( y = f(x) \) is the set of all possible \( y \) - values. So the range is \( \{y|y\in\mathbb{R}\} \) (or equivalently, \( \{y|-\infty

Answer:

\( \{y|y\in\mathbb{R}\} \) (the second option in the given choices)