1. **State the problem:** We are given the function $g(x) = 3x^2 + \frac{5}{x^3}$ and asked to analyze it on the interval $[-2, 2]$. 2. **Understand the function:** This function combines a quadratic term $3x^2$ and a rational term $\frac{5}{x^3}$. Note that $g(x)$ is undefined at $x=0$ because division by zero is not allowed. 3. **Domain consideration:** Since $g(x)$ has $\frac{5}{x^3}$, the domain excludes $x=0$. On the interval $[-2, 2]$, the function is defined for all $x$ except $0$. 4. **Find critical points:** To find critical points, compute the derivative: $$g'(x) = \frac{d}{dx}\left(3x^2 + \frac{5}{x^3}\right) = 6x - 15x^{-4} = 6x - \frac{15}{x^4}$$ 5. **Set derivative equal to zero to find critical points:** $$6x - \frac{15}{x^4} = 0$$ Multiply both sides by $x^4$ (noting $x \neq 0$): $$6x \cdot x^4 - 15 = 0 \Rightarrow 6x^5 - 15 = 0$$ 6. **Solve for $x$:** $$6x^5 = 15$$ $$x^5 = \frac{15}{6} = \frac{5}{2}$$ $$x = \sqrt[5]{\frac{5}{2}}$$ 7. **Check if this critical point lies in the interval $[-2, 2]$:** Since $\sqrt[5]{2.5} \approx 1.22$, it is within the interval. 8. **Evaluate $g(x)$ at critical point and endpoints:** - At $x = -2$: $$g(-2) = 3(-2)^2 + \frac{5}{(-2)^3} = 3 \cdot 4 + \frac{5}{-8} = 12 - \frac{5}{8} = 12 - 0.625 = 11.375$$ - At $x = 2$: $$g(2) = 3(2)^2 + \frac{5}{2^3} = 3 \cdot 4 + \frac{5}{8} = 12 + 0.625 = 12.625$$ - At $x = \sqrt[5]{\frac{5}{2}}$: $$g\left(\sqrt[5]{\frac{5}{2}}\right) = 3\left(\sqrt[5]{\frac{5}{2}}\right)^2 + \frac{5}{\left(\sqrt[5]{\frac{5}{2}}\right)^3} = 3 \left(\sqrt[5]{\left(\frac{5}{2}\right)^2}\right) + 5 \cdot \sqrt[5]{\left(\frac{5}{2}\right)^{-3}}$$ Simplify exponents: $$= 3 \left(\left(\frac{5}{2}\right)^{\frac{2}{5}}\right) + 5 \left(\left(\frac{5}{2}\right)^{-\frac{3}{5}}\right) = 3 \left(\frac{5}{2}\right)^{\frac{2}{5}} + 5 \left(\frac{5}{2}\right)^{-\frac{3}{5}}$$ This is the exact value; approximate numerically if needed. 9. **Summary:** The function has a critical point at $x = \sqrt[5]{\frac{5}{2}} \approx 1.22$ within the interval $[-2, 2]$, and is undefined at $x=0$. The values at the endpoints and critical point help understand the behavior of $g(x)$ on this interval. **Final answer:** Critical point at $x = \sqrt[5]{\frac{5}{2}}$ with function value $g\left(\sqrt[5]{\frac{5}{2}}\right) = 3 \left(\frac{5}{2}\right)^{\frac{2}{5}} + 5 \left(\frac{5}{2}\right)^{-\frac{3}{5}}$; function undefined at $x=0$; endpoint values $g(-2) = 11.375$, $g(2) = 12.625$.