1. **Stating the problem:** We have a function $f$ defined on the interval $[a,b]$ such that $f(x) \leq 0$ for all $x \in [a,b]$. We want to determine which of the given functions $g(x)$ 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 check if $g$ is increasing on $(a,b)$, we examine its derivative $g'(x)$. If $g'(x) > 0$ for all $x \in (a,b)$, then $g$ is strictly increasing there. 4. **Analyze each option:** **(a) $g(x) = [f(x)]^2$** - Derivative: $$g'(x) = 2 f(x) f'(x)$$ - Since $f(x) \leq 0$, $f(x)$ is non-positive. - The sign of $g'(x)$ depends on $f(x)$ and $f'(x)$. - If $f'(x)$ is positive, $g'(x)$ is negative or zero (because $f(x) \leq 0$), so $g$ is not necessarily increasing. **(b) $g(x) = x f(x)$** - Derivative: $$g'(x) = f(x) + x f'(x)$$ - Since $f(x) \leq 0$, the first term is non-positive. - Without more info on $f'(x)$, we cannot guarantee $g'(x) > 0$. **(c) $g(x) = [f(x)]^3$** - Derivative: $$g'(x) = 3 [f(x)]^2 f'(x)$$ - Since $[f(x)]^2 \geq 0$, the sign of $g'(x)$ depends on $f'(x)$. - If $f'(x) > 0$, then $g'(x) > 0$. - Because $f$ maps into $\mathbb{R}^-$ and the curve is decreasing downward, $f$ is increasing (less negative) if $f'(x) > 0$. - So $g$ is increasing if $f'(x) > 0$. **(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:** Among the options, only (c) $g(x) = [f(x)]^3$ is increasing on $(a,b)$ if $f'(x) > 0$, which aligns with the function $f$ being increasing (less negative) on that interval. **Final answer:** (c) $g(x) = [f(x)]^3$ is increasing on $(a,b)$.