\begin{tabular}{|c|c|}\hline\( g(x)=\frac{1}{5 x^{3}} \) & \( h(x)=x^{2}|x|+4 \) \\ Even & Even \\ Odd & Odd \\ Neither & Neither \\ \hline\end{tabular}

Function by following steps: - step0: Determine if even, odd or neither: \(h\left(x\right)=x^{2}\left|x\right|+4\) - step1: Use substitution: \(h\left(-x\right)=\left(-x\right)^{2}\left|-x\right|+4\) - step2: Simplify: \(h\left(-x\right)=x^{2}\left|x\right|+4\) - step3: Use substitution: \(h\left(-x\right)=h\left(x\right)\) - step4: The function is an even function: \(\textrm{Even}\) Determine whether the function \( g(x)=\frac{1}{5 x^{3}} \) is even or odd Function by following steps: - step0: Determine if even, odd or neither: \(g\left(x\right)=\frac{1}{5x^{3}}\) - step1: Use substitution: \(g\left(-x\right)=\frac{1}{5\left(-x\right)^{3}}\) - step2: Simplify: \(g\left(-x\right)=-\frac{1}{5x^{3}}\) - step3: Use substitution: \(g\left(-x\right)=-g\left(x\right)\) - step4: The function is an odd function: \(\textrm{Odd}\) Let's analyze the functions \( g(x) \) and \( h(x) \) to determine their properties regarding being even, odd, or neither. 1. **Function \( g(x) = \frac{1}{5 x^{3}} \)**: - After analysis, we find that \( g(x) \) is **Odd**. This means that \( g(-x) = -g(x) \) for all \( x \) in the domain of \( g \). 2. **Function \( h(x) = x^{2}|x| + 4 \)**: - After analysis, we find that \( h(x) \) is **Even**. This means that \( h(-x) = h(x) \) for all \( x \) in the domain of \( h \). ### Summary: - \( g(x) = \frac{1}{5 x^{3}} \) is **Odd**. - \( h(x) = x^{2}|x| + 4 \) is **Even**.