1. **Problem Statement:** We are given a loss function $$L(w) = w^4 - 4w^2 + 5$$ and need to find its critical points, classify them as local minima or maxima using the first derivative test, and explain the meaning of a minimum in model training. 2. **Find critical points:** Critical points occur where the first derivative is zero or undefined. Since $$L(w)$$ is a polynomial, the derivative exists everywhere. Compute the first derivative: $$L'(w) = \frac{d}{dw}(w^4 - 4w^2 + 5) = 4w^3 - 8w$$ Set the derivative equal to zero to find critical points: $$4w^3 - 8w = 0$$ Factor out $$4w$$: $$4w(w^2 - 2) = 0$$ So critical points satisfy: $$4w = 0 \implies w = 0$$ or $$w^2 - 2 = 0 \implies w = \pm \sqrt{2}$$ 3. **Classify critical points using the first derivative test:** - For $$w = 0$$: - Choose test points around 0, e.g., $$w = -1$$ and $$w = 1$$. - Evaluate $$L'(w)$$: - $$L'(-1) = 4(-1)^3 - 8(-1) = -4 + 8 = 4 > 0$$ (positive) - $$L'(1) = 4(1)^3 - 8(1) = 4 - 8 = -4 < 0$$ (negative) - Derivative changes from positive to negative, so $$w=0$$ is a local maximum. - For $$w = \sqrt{2}$$: - Test points: $$w=1.4$$ and $$w=1.5$$ (since $$\sqrt{2} \approx 1.414$$) - Evaluate $$L'(w)$$: - $$L'(1.4) = 4(1.4)^3 - 8(1.4) \approx 4(2.744) - 11.2 = 10.976 - 11.2 = -0.224 < 0$$ (negative) - $$L'(1.5) = 4(1.5)^3 - 8(1.5) = 4(3.375) - 12 = 13.5 - 12 = 1.5 > 0$$ (positive) - Derivative changes from negative to positive, so $$w=\sqrt{2}$$ is a local minimum. - For $$w = -\sqrt{2}$$: - Test points: $$w=-1.5$$ and $$w=-1.4$$ - Evaluate $$L'(w)$$: - $$L'(-1.5) = 4(-1.5)^3 - 8(-1.5) = 4(-3.375) + 12 = -13.5 + 12 = -1.5 < 0$$ (negative) - $$L'(-1.4) = 4(-1.4)^3 - 8(-1.4) \approx 4(-2.744) + 11.2 = -10.976 + 11.2 = 0.224 > 0$$ (positive) - Derivative changes from negative to positive, so $$w=-\sqrt{2}$$ is a local minimum. 4. **Meaning of a minimum in model training:** A minimum of the loss function corresponds to a set of model parameters $$w$$ where the loss is locally the smallest. This means the model predictions are closest to the true values in that neighborhood, indicating better performance. Training aims to find such minima to optimize the model's accuracy and generalization. **Final answers:** - Critical points: $$w = 0, \pm \sqrt{2}$$ - Local maxima: $$w=0$$ - Local minima: $$w=\pm \sqrt{2}$$