Answer
$1 + \alpha(x-1) + \beta(y-1) + \gamma(z-1) $
Work Step by Step
Step 1 We need to find the local linear approximation of each function at $(1,1,1)$ for the function given: \[ x^\alpha y^\beta z^\gamma \] Step 2 We can solve for the linear approximation of the function $f(x,y,z)$ at $(x_0, y_0, z_0)$ by using the formula below: \[ L(x,y,z) = f(x_0, y_0, z_0) + f_x(x_0, y_0, z_0)(x-x_0) + f_y(x_0, y_0, z_0)(y-y_0) + f_z(x_0, y_0, z_0)(z-z_0) \] Step 3 Solving for $f(x_0, y_0, z_0)$: \[ f(x_0, y_0, z_0) = x^\alpha y^\beta z^\gamma = (1)(1)(1) = 1 \] Step 4 Solving for $f_x(x_0, y_0, z_0)$: \[ f_x(x_0, y_0, z_0) = \frac{\partial}{\partial x}(x^\alpha y^\beta z^\gamma) = y^\beta z^\gamma \frac{\partial}{\partial x}(x^\alpha) = y^\beta z^\gamma \alpha x^{\alpha-1} = \alpha x^{\alpha-1} y^\beta z^\gamma = \alpha \] Step 5 Solving for $f_y(x_0, y_0, z_0)$: \[ f_y(x_0, y_0, z_0) = \frac{\partial}{\partial y}(x^\alpha y^\beta z^\gamma) = x^\alpha z^\gamma \frac{\partial}{\partial y}(y^\beta) = x^\alpha z^\gamma \beta y^{\beta-1} = \beta x^\alpha y^{\beta-1} z^\gamma = \beta \] Step 6 Solving for $f_z(x_0, y_0, z_0)$: \[ f_z(x_0, y_0, z_0) = \frac{\partial}{\partial z}(x^\alpha y^\beta z^\gamma) = x^\alpha y^\beta \frac{\partial}{\partial z}(z^\gamma) = x^\alpha y^\beta \gamma z^{\gamma-1} = \gamma x^\alpha y^\beta z^{\gamma-1} = \gamma \] Now that we have calculated each component, we can substitute this back into the formula at $(x_0, y_0, z_0) = (1,1,1)$: \[ \begin{aligned} L(x, y, z) &= L(x_0, y_0, z_0) + \frac{\partial L}{\partial x}(x_0, y_0, z_0) (x - x_0) \\ &\quad + \frac{\partial L}{\partial y}(x_0, y_0, z_0) (y - y_0) + \frac{\partial L}{\partial z}(x_0, y_0, z_0) (z - z_0) \\ &= 1 + \alpha (x - x_0) + \beta (y - y_0) + \gamma (z - z_0) \\ &= 1 + \alpha (x - 1) + \beta (y - 1) + \gamma (z - 1) \end{aligned} \] Thus, \[ x^\alpha y^\beta z^\gamma \approx 1 + \alpha(x-1) + \beta(y-1) + \gamma(z-1) \] Result \[ 1 + \alpha(x-1) + \beta(y-1) + \gamma(z-1) \]
