Let's solve each part (a, b, c, d) in **2 steps** with explanations and the **final answer**. --- ### **a) \(xy + z - \cos(x+y+z) + 1 = 0\)** #### **Step 1: Check if the equation can be solved for \(z\) near \((0,0,0)\).** Let \(F(x, y, z) = xy + z - \cos(x+y+z) + 1\). Check \(F(0,0,0) = 0 \cdot 0 + 0 - \cos(0+0+0) + 1 = 0 - 1 + 1 = 0\). Now, check \(\frac{\partial F}{\partial z}\) at \((0,0,0)\): \[ \frac{\partial F}{\partial z} = 1 + \sin(x+y+z) \] At \((0,0,0)\), this gives \(1 + 0 = 1 \neq 0\). **Explanation:** By the **Implicit Function Theorem**, since \(F(0,0,0) = 0\) and \(\frac{\partial F}{\partial z}(0,0,0) \neq 0\), there exists a differentiable function \(z(x, y)\) near \((0, 0)\). --- #### **Step 2: Is the solution differentiable near \((0,0)\)?** Yes, since \(F\) is smooth and \(\frac{\partial F}{\partial z} \neq 0\), the solution is differentiable. --- **Final answer (a):** **Yes, the equation can be solved for \(z\) near \((0,0)\), and the solution is differentiable near \((0,0)\).** --- ### **b) \(x + y^2 + z^3 + \sqrt{\cos(x^2+y^2)+4z+8} = 3\)** #### **Step 1: Check solvability at \((0,0,0)\).** Let \(F(x, y, z) = x + y^2 + z^3 + \sqrt{\cos(x^2 + y^2) + 4z + 8} - 3\). Plug in \((0,0,0)\): \[ F(0,0,0) = 0 + 0 + 0 + \sqrt{\cos(0) + 0 + 8} - 3 = \sqrt{1 + 8} - 3 = 3 - 3 = 0 \] Now, check \(\frac{\partial F}{\partial z}\) at \((0,0,0)\): \[ \frac{\partial F}{\partial z} = 3z^2 + \frac{4}{2\sqrt{\cos(x^2+y^2)+4z+8}} \] At \((0,0,0)\), this is \(0 + \frac{4}{2 \cdot 3} = \frac{2}{3} \neq 0\). **Explanation:** By the Implicit Function Theorem, since \(F(0,0,0) = 0\) and \(\frac{\partial F}{\partial z}(0,0,0) \neq 0\), there exists a differentiable solution for \(z\) near \((0,0)\). --- #### **Step 2: Is the solution differentiable near \((0,0)\)?** Yes, as all functions involved are smooth and the partial derivative is nonzero. --- **Final answer (b):** **Yes, the equation can be solved for \(z\) near \((0,0)\), and the solution is differentiable near \((0,0)\).** --- ### **c) \(x(2\sin y - 3y \sin z) + (z \sin x - \cos z) = 0\)** #### **Step 1: Check solvability at \((0,0,0)\).** Let \(F(x, y, z) = x(2\sin y - 3y \sin z) + (z \sin x - \cos z)\). Plug in \((0,0,0)\): \[ F(0,0,0) = 0 + (0 - 1) = -1 \neq 0 \] **Explanation:** Since \(F(0,0,0) \neq 0\), the equation **cannot** be solved for \(z\) near \((0,0)\). --- #### **Step 2: Is the solution differentiable near \((0,0)\)?** No solution exists near \((0,0)\), so differentiability is not relevant. --- **Final answer (c):** **No, the equation cannot be solved for \(z\) near \((0,0)\).** --- ### **d) \(x^2 + y^2 + z^2 + g(x, y, z) = 0\), where \(g(0,0,0) = 0\), \(g_z(0,0,0) > 0\)** #### **Step 1: Check solvability at \((0,0,0)\).** Let \(F(x, y, z) = x^2 + y^2 + z^2 + g(x, y, z)\). Plug in \((0,0,0)\): \[ F(0,0,0) = 0 + 0 + 0 + 0 = 0 \] \[ \frac{\partial F}{\partial z} = 2z + g_z(x, y, z) \] At \((0,0,0)\): \(0 + g_z(0,0,0) > 0\). **Explanation:** By the Implicit Function Theorem, a differentiable solution for \(z\) exists near \((0,0)\). --- #### **Step 2: Is the solution differentiable near \((0,0)\)?** Yes, since \(g\) is continuously differentiable and \(\frac{\partial F}{\partial z}(0,0,0) > 0\). --- **Final answer (d):** **Yes, the equation can be solved for \(z\) near \((0,0)\), and the solution is differentiable near \((0,0)\).** --- ## **Summary Table** | Part | Solvable for \(z\) near (0,0,0)? | Differentiable? | |------|:-------------------------------:|:--------------:| | a | Yes | Yes | | b | Yes | Yes | | c | No | Not applicable | | d | Yes | Yes |