1. **Problem statement:** We have a function $f$ defined on the interval $[a,b]$ with values in $\mathbb{R}^-$ (negative real numbers). We want to find which function $g$ among the options is increasing on the open interval $(a,b)$. 2. **Given options:** (a) $g(x) = [f(x)]^2$ (b) $g(x) = x \times f(x)$ (c) $g(x) = [f(x)]^3$ (d) $g(x) = -2x - f(x)$ 3. **Key idea:** To determine if $g$ is increasing on $(a,b)$, we check if its derivative $g'(x) > 0$ for all $x \in (a,b)$. 4. **Analyze each option:** - (a) $g(x) = [f(x)]^2$ Derivative: $$g'(x) = 2 f(x) f'(x)$$ Since $f(x) < 0$ on $[a,b]$, $f(x)$ is negative. The sign of $g'(x)$ depends on $f'(x)$. If $f$ is increasing, $f'(x) > 0$, then $g'(x) = 2 \times (\text{negative}) \times (\text{positive}) < 0$, so $g$ is decreasing, not increasing. - (b) $g(x) = x f(x)$ Derivative: $$g'(x) = f(x) + x f'(x)$$ Since $f(x) < 0$, the first term is negative. The second term depends on $f'(x)$ and $x$. Without more info, we cannot 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 on $f'(x)$. If $f$ is increasing ($f'(x) > 0$), then $g'(x) > 0$, so $g$ is increasing. - (d) $g(x) = -2x - f(x)$ Derivative: $$g'(x) = -2 - f'(x)$$ For $g'(x) > 0$, we need $-2 - f'(x) > 0 \Rightarrow f'(x) < -2$. This is a strong condition and not guaranteed. 5. **Conclusion:** The function $g(x) = [f(x)]^3$ is increasing on $(a,b)$ if $f$ is increasing, which matches the graph description where $f$ increases from $a$ to $b$. **Final answer:** (c) $g(x) = [f(x)]^3$ is increasing on $(a,b)$.