Math Problem Statement

Given the function g(x, y, z) = z^3 + xz + y, apply the Intermediate Value Theorem, implicit differentiation, and linear approximation to solve for z and approximate the root of the cubic equation z^3 + 19z - 640 = 0.

Solution

Let's break down the problem step by step:

  1. We are given the function g(x,y,z)=z3+xz+yg(x, y, z) = z^3 + xz + y.
  2. We have limits as z+z \to +\infty and zz \to -\infty, showing that g(x,y,z)g(x, y, z) \to \infty and g(x,y,z)g(x, y, z) \to -\infty respectively.
  3. This suggests that by the Intermediate Value Theorem, the equation g(x,y,z)=0g(x, y, z) = 0 has a solution for each xx and yy.

Differentiation Process:

  • We are asked to find the linear approximation of f(x,y)f(x, y) at (14,624)(14, -624) using implicit differentiation. This means we treat zz as a function of xx and yy and differentiate both sides of the equation g(x,y,z)=0g(x, y, z) = 0.

    First, we differentiate implicitly: gx=ddx(z3+xz+y)=0\frac{\partial g}{\partial x} = \frac{d}{dx}(z^3 + xz + y) = 0 Applying the chain rule: 3z2dzdx+z+xdzdx=03z^2 \frac{dz}{dx} + z + x \frac{dz}{dx} = 0 Solving for dzdx\frac{dz}{dx}: dzdx=z3z2+x\frac{dz}{dx} = \frac{-z}{3z^2 + x}

    Similarly, for gy\frac{\partial g}{\partial y}: ddy(z3+xz+y)=0\frac{d}{dy}(z^3 + xz + y) = 0 Since the only term involving yy is linear: dzdy=13z2+x\frac{dz}{dy} = \frac{-1}{3z^2 + x}

  1. Given that g(14,624,8)=0g(14, -624, 8) = 0, we can use these values for x=14x = 14, y=624y = -624, and z=8z = 8 to evaluate the derivatives:

    dzdx=83(8)2+14=83(64)+14=8192+14=82060.0388\frac{dz}{dx} = \frac{-8}{3(8)^2 + 14} = \frac{-8}{3(64) + 14} = \frac{-8}{192 + 14} = \frac{-8}{206} \approx -0.0388 dzdy=13(8)2+14=12060.00485\frac{dz}{dy} = \frac{-1}{3(8)^2 + 14} = \frac{-1}{206} \approx -0.00485

  2. Using the point (14,624)(14, -624), the linear approximation formula is: f(x,y)f(14,624)+dzdx(x14)+dzdy(y+624)f(x, y) \approx f(14, -624) + \frac{dz}{dx} (x - 14) + \frac{dz}{dy} (y + 624) Since f(14,624)=8f(14, -624) = 8, the linear approximation becomes: f(x,y)80.0388(x14)0.00485(y+624)f(x, y) \approx 8 - 0.0388(x - 14) - 0.00485(y + 624)

Root Approximation:

Finally, you're asked to use this approximation to estimate the root of the equation z3+19z640=0z^3 + 19z - 640 = 0. This can be done by solving iteratively, but the first approximation would be based on f(x,y)8f(x, y) \approx 8 as a starting point.


Do you want more details or clarification on any of the steps? Here are five relative questions for further exploration:

  1. How does the Intermediate Value Theorem apply in multivariable contexts like this one?
  2. Why is implicit differentiation necessary in this problem?
  3. Can you find the exact root of z3+19z640=0z^3 + 19z - 640 = 0 using the approximation?
  4. What other approximation methods could be used to solve the cubic equation?
  5. How do changes in xx and yy influence the solution zz in g(x,y,z)=0g(x, y, z) = 0?

Tip: When performing implicit differentiation, always apply the chain rule carefully to each variable that depends on the others.

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Math Problem Analysis

Mathematical Concepts

Algebra
Multivariable Functions
Implicit Differentiation
Linear Approximation
Intermediate Value Theorem

Formulas

g(x, y, z) = z^3 + xz + y
Implicit differentiation: 3z^2 (dz/dx) + z + x (dz/dx) = 0
Linear approximation formula: f(x, y) ≈ f(14, -624) + (dz/dx) (x - 14) + (dz/dy) (y + 624)

Theorems

Intermediate Value Theorem
Implicit Differentiation

Suitable Grade Level

Grades 11-12, College Level