1. **State the problem:** We need to find the critical points of the function $$g(t) = t^2 \sqrt[3]{2t} - 5$$. Critical points occur where the derivative $$g'(t)$$ is zero or undefined. 2. **Rewrite the function:** Express the cube root as a fractional exponent: $$g(t) = t^2 (2t)^{1/3} - 5 = t^2 \cdot 2^{1/3} t^{1/3} - 5 = 2^{1/3} t^{2 + \frac{1}{3}} - 5 = 2^{1/3} t^{\frac{7}{3}} - 5$$ 3. **Find the derivative:** Using the power rule $$\frac{d}{dt} t^n = n t^{n-1}$$, $$g'(t) = 2^{1/3} \cdot \frac{7}{3} t^{\frac{7}{3} - 1} = 2^{1/3} \cdot \frac{7}{3} t^{\frac{4}{3}}$$ 4. **Set the derivative equal to zero to find critical points:** $$g'(t) = 0 \implies 2^{1/3} \cdot \frac{7}{3} t^{\frac{4}{3}} = 0$$ Since $$2^{1/3}$$ and $$\frac{7}{3}$$ are nonzero constants, the equation reduces to: $$t^{\frac{4}{3}} = 0 \implies t = 0$$ 5. **Check where the derivative is undefined:** The derivative involves $$t^{4/3}$$ which is defined for all real $$t$$ (including zero), so no points where $$g'(t)$$ is undefined. 6. **Conclusion:** The only critical point is at $$t = 0$$. **Final answer:** The critical point of $$g(t)$$ is at $$t = 0$$.