1. **Problem Statement:** We analyze the function $$f(x) = 3 \ln(x + 2) - x$$ to find where it is concave up and concave down, and explain why concave-up regions help gradient descent convergence. 2. **Recall:** - The concavity of a function is determined by the sign of its second derivative $$f''(x)$$. - If $$f''(x) > 0$$, the function is concave up on that interval. - If $$f''(x) < 0$$, the function is concave down on that interval. 3. **First derivative:** $$f'(x) = \frac{d}{dx} \left(3 \ln(x+2) - x\right) = 3 \cdot \frac{1}{x+2} - 1 = \frac{3}{x+2} - 1$$ 4. **Second derivative:** $$f''(x) = \frac{d}{dx} \left(\frac{3}{x+2} - 1\right) = -3 \cdot \frac{1}{(x+2)^2} = -\frac{3}{(x+2)^2}$$ 5. **Analyze concavity:** - Since $$ (x+2)^2 > 0 $$ for all $$x \neq -2$$, and the numerator is $$-3$$ (negative), we have $$f''(x) = -\frac{3}{(x+2)^2} < 0$$ for all $$x > -2$$ (domain of $$f$$). - Therefore, $$f(x)$$ is concave down on its entire domain $$(-2, \infty)$$. - There are no intervals where $$f(x)$$ is concave up. 6. **Why concave-up regions help gradient descent:** - Gradient descent converges faster in concave-up regions because the function curves upward, making the gradient point more directly toward the minimum. - In concave-up regions, the second derivative is positive, indicating the function is locally convex, which ensures a unique local minimum and stable descent steps. - Concave-down regions can cause gradient descent to diverge or oscillate because the function curves downward, potentially leading to maxima or saddle points. **Final answers:** - (a) $$f(x)$$ is concave down on $$(-2, \infty)$$ and has no concave-up intervals. - (b) Concave-up regions help gradient descent converge because they indicate local convexity, ensuring stable and directed steps toward a minimum.