1. **State the problem:** We are given the function $$f(x) = -\frac{1}{3}x^3 + 3x^2 - 5x - 12$$ and a relative minimum point at $\left(1, \frac{43}{3}\right)$. We want to understand the behavior of the function at this point, including concavity. 2. **Recall formulas:** - The first derivative $$f'(x)$$ gives the slope of the tangent line and helps find critical points. - The second derivative $$f''(x)$$ tells us about concavity: if $$f''(x) > 0$$, the graph is concave up; if $$f''(x) < 0$$, concave down. 3. **Find the first derivative:** $$f'(x) = -\frac{1}{3} \cdot 3x^2 + 3 \cdot 2x - 5 = -x^2 + 6x - 5$$ 4. **Verify the critical point at $x=1$:** $$f'(1) = -(1)^2 + 6(1) - 5 = -1 + 6 - 5 = 0$$ This confirms $x=1$ is a critical point. 5. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(-x^2 + 6x - 5) = -2x + 6$$ 6. **Evaluate the second derivative at $x=1$:** $$f''(1) = -2(1) + 6 = 4$$ Since $$f''(1) = 4 > 0$$, the graph is concave up at $x=1$, confirming a relative minimum. 7. **Check the function value at $x=1$ to confirm the point:** $$f(1) = -\frac{1}{3}(1)^3 + 3(1)^2 - 5(1) - 12 = -\frac{1}{3} + 3 - 5 - 12 = -\frac{1}{3} - 14 = -\frac{43}{3}$$ Note: The given point is $\left(1, \frac{43}{3}\right)$ but calculation shows $-\frac{43}{3}$. This suggests the correct point is $\left(1, -\frac{43}{3}\right)$. **Final answer:** The function has a relative minimum at $$\left(1, -\frac{43}{3}\right)$$ where the graph is concave up since $$f''(1) = 4 > 0$$.