Problem 13 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}-y^{2}+4 x-4 y-8 $$

Step-by-Step Solution

Answer

The function \(f(x, y)=x^{2}-y^{2}+4 x-4 y-8\) has a saddle point at \((-2, -2)\).

Step 1: Find the First Partial Derivatives

For the function \(f(x, y) = x^{2} - y^{2} + 4x - 4y - 8\), the first partial derivatives are found through differentiation of the function \(f\) with respect to each variable while keeping the other constant. The derivative with respect to \(x\) is denoted \(f_x\) and the derivative with respect to \(y\) is denoted \(f_y\). Performing this yields: \[f_x(x, y) = 2x + 4\] and \[f_y(x, y) = -2y - 4\].

Step 2: Locate Critical Points

Critical points occur where the gradient of a function is zero (or undefined). As only real number solutions are considered, the point will be a critical point if both the first partial derivatives \(f_x\) and \(f_y\) equal zero. So we solve the following system of equations: \[2x + 4 = 0\] \[-2y - 4 = 0\]. Doing so gives us a critical point at \((-2, -2)\).

Step 3: Calculate the Second Partial Derivatives

These second derivatives show how the first derivative changes, giving information about the curvature of the function. The second partial derivatives are calculated in the same way as the first, resulting in: \[f_{xx}(x, y) = 2\] \[f_{yy}(x, y) = -2\] \[f_{xy}(x, y) = f_{yx}(x,y) = 0\].

Step 4: Apply the Second Derivative Test

The Second Derivative Test utilizes a determinant of a matrix (called the Hessian Matrix) made up of the second partial derivatives evaluated at the critical point. If the determinant is greater than zero, then the function has a local maximum or minimum at that point. If it's less than zero, the function has a saddle point. If it's equal to zero, the test is inconclusive. Here, the Hessian matrix \(\begin{bmatrix}2&0\0&-2\end{bmatrix}\) gives the determinant as \(-4\). As this is less than zero, \((-2, -2)\) is a saddle point.