1. **Problem statement:** We have a function $f$ defined on the interval $[a,b]$ with values in $\mathbb{R}^-$ (meaning $f(x) < 0$ for all $x \in [a,b]$). We want to determine for which function $g$ among the options (a) $[f(x)]^2$, (b) $x \times f(x)$, (c) $[f(x)]^3$, and (d) $-2x - f(x)$ the function $g$ is increasing on the open interval $(a,b)$. 2. **Key idea:** To check if $g$ is increasing on $(a,b)$, we need to check if its derivative $g'(x) > 0$ for all $x \in (a,b)$. 3. **Analyze each option:** - (a) $g(x) = [f(x)]^2$. Derivative: $$g'(x) = 2 f(x) f'(x)$$ Since $f(x) < 0$, the sign of $g'(x)$ depends on $f'(x)$ and $f(x)$. Without knowing $f'(x)$, we cannot guarantee $g'(x) > 0$. - (b) $g(x) = x f(x)$. Derivative: $$g'(x) = f(x) + x f'(x)$$ Again, sign depends on $f(x)$ and $f'(x)$; no guarantee $g'(x) > 0$. - (c) $g(x) = [f(x)]^3$. Derivative: $$g'(x) = 3 [f(x)]^2 f'(x)$$ Since $[f(x)]^2 > 0$ always, the sign of $g'(x)$ depends solely on $f'(x)$. - (d) $g(x) = -2x - f(x)$. Derivative: $$g'(x) = -2 - f'(x)$$ For $g'(x) > 0$, we need $-2 - f'(x) > 0 \Rightarrow f'(x) < -2$. 4. **Given $f$ maps to negative values, and the problem context suggests $f$ is decreasing (since the curve dips below x-axis and then rises), the safest choice for guaranteed increasing $g$ is (c) $[f(x)]^3$ if $f'(x) > 0$ (i.e., $f$ increasing). But since $f$ is negative, $[f(x)]^3$ preserves the sign and the derivative depends on $f'(x)$. 5. **Conclusion:** Among the options, only (c) $[f(x)]^3$ is increasing on $(a,b)$ if $f$ is increasing (i.e., $f'(x) > 0$). This matches the problem's condition that $g$ is increasing on $(a,b)$. **Final answer:** (c) $g(x) = [f(x)]^3$ is increasing on the interval $(a,b)$.