Question

n identify its domain

1. $f(x) = \sqrt3{x + 2}$

Explanation:

Step1: Recall domain rules for cube roots

The cube root function \( \sqrt[3]{u} \) is defined for all real numbers \( u \), because we can take the cube root of any real number (positive, negative, or zero).

Step2: Analyze the expression inside the cube root

In the function \( f(x)=\sqrt[3]{x + 2} \), the expression inside the cube root is \( x+2 \). Since \( x+2 \) is a linear expression (a polynomial of degree 1) and polynomials are defined for all real numbers \( x \), and the cube root of any real number is defined, there are no restrictions on the value of \( x \).

Answer:

The domain of \( f(x)=\sqrt[3]{x + 2} \) is all real numbers, which can be written in interval notation as \( (-\infty, \infty) \) or in set - builder notation as \( \{x|x\in\mathbb{R}\} \).