1. **Problem statement:** Given that $f$ is decreasing on its domain, determine which of the following functions must be increasing on the same domain: (a) $y = -f(x)$ (b) $y = [f(x)]^3$ (c) $y = \ln(f(x))$ (d) $y = 2f(x)$ 2. **Recall definition:** A function $g$ is increasing if for any $x_1 < x_2$ in its domain, $g(x_1) \le g(x_2)$. 3. **Analyze each function:** (a) $y = -f(x)$: Since $f$ is decreasing, for $x_1 < x_2$, $f(x_1) \ge f(x_2)$. Multiplying by $-1$ reverses inequality: $$-f(x_1) \le -f(x_2)$$ So $y=-f(x)$ is increasing. (b) $y = [f(x)]^3$: Since the cube function $z \mapsto z^3$ is strictly increasing for all real $z$, composing a decreasing $f$ with an increasing cube function yields a decreasing function. Hence, $[f(x)]^3$ remains decreasing, not increasing. (c) $y = \ln(f(x))$: Logarithm $\ln(z)$ is increasing for $z > 0$. For $\ln(f(x))$ to be defined, $f(x) > 0$. Because $f$ is decreasing, $f(x_1) \ge f(x_2)$ for $x_1 < x_2$. Then: $$\ln(f(x_1)) \ge \ln(f(x_2))$$ Thus $\ln(f(x))$ is decreasing (since order preserved by $\ln$), not increasing. (d) $y = 2f(x)$: Multiplying a decreasing function by positive scalar $2$ retains decreasing nature. Thus $2f(x)$ is decreasing, not increasing. 4. **Conclusion:** Only function (a) $y=-f(x)$ is increasing on the domain. **Final answer:** (a)