1. **Problem statement:** Determine the sign of the second derivative at points A, B, C, and D for the given functions. 2. **Function a:** $y = x^3 - 6x^2 - 15x + 10$ - First derivative: $$y' = 3x^2 - 12x - 15$$ - Second derivative: $$y'' = 6x - 12$$ 3. **Evaluate $y''$ at points A, B, C, D:** - Since points are on the graph, we estimate their $x$-coordinates: - Point A: bottom-right quadrant, increasing function, approximate $x > 2$ - Point B: bottom-right quadrant, decreasing function, approximate $x$ near 3 - Point C: on vertical line at $x=0$ - Point D: top-right quadrant, decreasing and flattening, approximate $x > 2$ Calculate: - At $x=0$ (Point C): $$y'' = 6(0) - 12 = -12 < 0$$ (negative) - At $x=3$ (approx for A, B, D): $$y'' = 6(3) - 12 = 18 - 12 = 6 > 0$$ (positive) 4. **Points of inflection:** - Inflection points occur where $y''=0$: $$6x - 12 = 0 \implies x = 2$$ 5. **Function c:** $s = t + t^{-1}$ - First derivative: $$s' = 1 - t^{-2}$$ - Second derivative: $$s'' = 0 + 2t^{-3} = 2t^{-3} = \frac{2}{t^3}$$ 6. **Sign of $s''$ at points:** - For $t > 0$, $s'' > 0$ - For $t < 0$, $s'' < 0$ 7. **Points of inflection for $s$:** - Set $s''=0$: $$\frac{2}{t^3} = 0$$ - No solution since denominator cannot be zero and numerator is constant. - However, $s''$ is undefined at $t=0$, which is a vertical asymptote, not an inflection point. **Final answers:** - For $y$: $y''$ is negative at $x=0$ (Point C), positive at $x=3$ (Points A, B, D). - Inflection point for $y$ at $x=2$. - For $s$: $s''$ positive for $t>0$, negative for $t<0$, no inflection points.