1. **Problem Statement:** We have a neuron function defined as $f(x) = \text{ReLU}(wx + b)$ where $\text{ReLU}(z) = \max(0, z)$. Given parameters: $w = 2.5$, $b = -1$. We need to find the differential $df$ at $x=1$ and $x=0.3$, analyze where the differential is discontinuous, and explain the "dying ReLU" problem using differential analysis. 2. **Formula and Important Rules:** The ReLU function is defined as: $$\text{ReLU}(z) = \begin{cases} z & \text{if } z > 0 \\ 0 & \text{if } z \leq 0 \end{cases}$$ The differential $df$ of $f(x)$ at a point $x$ is given by: $$df = f'(x) dx$$ where $f'(x)$ is the derivative of $f$ with respect to $x$. Since $f(x) = \text{ReLU}(wx + b)$, by chain rule: $$f'(x) = \text{ReLU}'(wx + b) \cdot w$$ The derivative of ReLU is: $$\text{ReLU}'(z) = \begin{cases} 1 & \text{if } z > 0 \\ 0 & \text{if } z < 0 \end{cases}$$ Note: At $z=0$, the derivative is undefined (discontinuous). 3. **Calculate $df$ at $x=1$:** Calculate $z = wx + b = 2.5 \times 1 - 1 = 1.5$. Since $z=1.5 > 0$, $$f'(1) = 1 \times 2.5 = 2.5$$ Therefore, $$df = 2.5 \, dx$$ 4. **Calculate $df$ at $x=0.3$:** Calculate $z = 2.5 \times 0.3 - 1 = 0.75 - 1 = -0.25$. Since $z = -0.25 < 0$, $$f'(0.3) = 0 \times 2.5 = 0$$ Therefore, $$df = 0 \, dx$$ 5. **Analyze where the differential is discontinuous:** The differential is discontinuous where $z = wx + b = 0$. Solve for $x$: $$2.5x - 1 = 0 \implies x = \frac{1}{2.5} = 0.4$$ At $x=0.4$, the derivative changes abruptly from 0 to 2.5, causing a discontinuity in the differential. 6. **Implications of the discontinuity:** At $x=0.4$, the function transitions from inactive (output zero) to active (positive output). This sharp change means small changes in $x$ near 0.4 can cause sudden changes in the gradient, which affects learning dynamics. 7. **Explain the "dying ReLU" problem using differential analysis:** The "dying ReLU" problem occurs when the neuron output is zero for many inputs, i.e., when $wx + b \leq 0$. In this region, the derivative $f'(x) = 0$, so the differential $df = 0$. This means the neuron stops learning because gradients are zero and weights do not update. This can cause neurons to become inactive permanently, "dying" during training. **Final answers:** (a) $df = 2.5 \, dx$ at $x=1$ (b) $df = 0$ at $x=0.3$ (c) Differential is discontinuous at $x=0.4$ (d) "Dying ReLU" happens when $f'(x) = 0$ causing zero gradients and no learning.