Question

Let \( p = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}) \in {R^4 \) and \( f : {R}^4 \to {R} \) be a differentiable function such that \( f(p) = 6 \) and \( f(Ax) = A^3 f(x) \), for every \( A \in (0, \infty) \) and \( x \in {R}^4 \). The value of \[ 12 \frac{\partial f}{\partial x_1}(p) + 6 \frac{\partial f}{\partial x_2}(p) + 4 \frac{\partial f}{\partial x_3}(p) + 3 \frac{\partial f}{\partial x_4}(p) \] is equal to (answer in integer):}

Hint:

Use the scaling property of \( f \) to derive relationships between partial derivatives.

Answer 1

Step 1: Scaling property of \( f \). The functional equation \( f(Ax) = A^3 f(x) \) implies a relationship between \( f \) and its partial derivatives: \[ x_1 \frac{\partial f}{\partial x_1} + x_2 \frac{\partial f}{\partial x_2} + x_3 \frac{\partial f}{\partial x_3} + x_4 \frac{\partial f}{\partial x_4} = 3f(x). \] Step 2: Substitution at \( p \). At \( p = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}) \), substitute into the scaling property and simplify: \[ 12 \frac{\partial f}{\partial x_1}(p) + 6 \frac{\partial f}{\partial x_2}(p) + 4 \frac{\partial f}{\partial x_3}(p) + 3 \frac{\partial f}{\partial x_4}(p) = 216. \] Step 3: Conclusion. The value is \( {216} \).