1. We are asked to find the maximum and minimum values of the functions $\max(-m, 2m^3)$ and $\min(m, -2m^3)$.\n\n2. The problem involves comparing two expressions for each function and determining which is greater (for max) or smaller (for min) depending on the value of $m$.\n\n3. For $\max(-m, 2m^3)$:\n- We compare $-m$ and $2m^3$.\n- The maximum is $-m$ when $-m > 2m^3$, and $2m^3$ otherwise.\n- Solve inequality $-m > 2m^3$:\n$$-m > 2m^3 \implies -m - 2m^3 > 0 \implies -m(1 + 2m^2) > 0.$$\n- Since $1 + 2m^2 > 0$ for all real $m$, the sign depends on $-m$, so:\n$$-m > 0 \implies m < 0.$$\n- Therefore, for $m < 0$, $\max(-m, 2m^3) = -m$, and for $m \geq 0$, $\max(-m, 2m^3) = 2m^3$.\n\n4. For $\min(m, -2m^3)$:\n- We compare $m$ and $-2m^3$.\n- The minimum is $m$ when $m < -2m^3$, and $-2m^3$ otherwise.\n- Solve inequality $m < -2m^3$:\n$$m + 2m^3 < 0 \implies m(1 + 2m^2) < 0.$$\n- Since $1 + 2m^2 > 0$ always, the inequality depends on $m$, so:\n$$m < 0.$$\n- Therefore, for $m < 0$, $\min(m, -2m^3) = m$, and for $m \geq 0$, $\min(m, -2m^3) = -2m^3$.\n\n5. Summary:\n$$\max(-m, 2m^3) = \begin{cases} -m & m < 0 \\ 2m^3 & m \geq 0 \end{cases}$$\n$$\min(m, -2m^3) = \begin{cases} m & m < 0 \\ -2m^3 & m \geq 0 \end{cases}$$