Question

what type of function is $g(x) = \sqrt3{x}$? (1 point)

\bigcirc neither even nor odd

\bigcirc odd

\bigcirc even

\bigcirc both even and odd

Explanation:

Step1: Recall the definitions

A function \( y = f(x) \) is even if \( f(-x)=f(x) \) for all \( x \) in the domain, and odd if \( f(-x)= - f(x) \) for all \( x \) in the domain. The domain of \( g(x)=\sqrt[3]{x} \) is all real numbers (\( \mathbb{R} \)).

Step2: Compute \( g(-x) \)

We have \( g(-x)=\sqrt[3]{-x} \). Using the property of cube roots \( \sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b} \), we can write \( \sqrt[3]{-x}=\sqrt[3]{-1\cdot x}=\sqrt[3]{-1}\cdot\sqrt[3]{x} \). Since \( \sqrt[3]{-1}=- 1 \), this simplifies to \( - \sqrt[3]{x} \). And we know that \( -g(x)=-\sqrt[3]{x} \). So \( g(-x)=-g(x) \).

Answer:

B. odd (assuming the options are labeled as: A. neither even nor odd, B. odd, C. even, D. both even and odd)