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Find all the local maxima, local minima, and saddle points of the functions in Exercises \(1-30 .\) $$ f(x, y)=x^{2}-y^{2}-2 x+4 y+6 $$

Short Answer

Expert verified
The critical point (1, 2) is a saddle point.

Step by step solution

01

Find the First Partial Derivatives

To find the critical points, we first need to compute the first partial derivatives of the function with respect to both variables. For the given function \( f(x, y) = x^2 - y^2 - 2x + 4y + 6 \), the first partial derivative with respect to \( x \) is:\[ f_x = \frac{\partial}{\partial x}(x^2 - y^2 - 2x + 4y + 6) = 2x - 2 \]The first partial derivative with respect to \( y \) is:\[ f_y = \frac{\partial}{\partial y}(x^2 - y^2 - 2x + 4y + 6) = -2y + 4 \]
02

Solve for Critical Points

Critical points occur where both \( f_x \) and \( f_y \) are equal to zero.- Set \( 2x - 2 = 0 \) to find \( x = 1 \).- Set \( -2y + 4 = 0 \) to find \( y = 2 \).Thus, the critical point is \((x, y) = (1, 2)\).
03

Compute Second Partial Derivatives

To determine the nature of the critical point, compute the second partial derivatives:- \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \)- \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2 \)- \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \)- \( f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = 0 \)
04

Determine the Nature of the Critical Point

Use the second derivative test with the determinant of the Hessian matrix, \( D = f_{xx}f_{yy} - (f_{xy})^2 \):\[ D = 2*(-2) - 0^2 = -4 \]Since \( D < 0 \), the critical point \((1, 2)\) is a saddle point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are a fundamental concept in calculus, helping us understand the behavior of multivariable functions. When dealing with a function of two variables like \( f(x, y) \), a partial derivative is the derivative of the function with respect to one variable while keeping the other constant.

In the context of the given exercise, the first partial derivative of \( f \) with respect to \( x \) is denoted \( f_x \), and the one with respect to \( y \) is \( f_y \). These derivatives help in locating critical points, which are potential candidates for local maxima, minima, or saddle points. By finding where both partial derivatives equal zero, you identify points where the function is flat, thereby revealing critical points.
  • \( f_x = 2x - 2 \)
  • \( f_y = -2y + 4 \)
These expressions allow us to solve for critical points by setting \( f_x = 0 \) and \( f_y = 0 \), leading to \( (x, y) = (1, 2) \). Understanding partial derivatives creates the foundation for exploring more complex features of a function's graph.
Second Derivative Test
The second derivative test is a powerful tool to determine the nature of critical points found in a multivariable function. Once you have the critical points, the second derivative test involves computing the second partial derivatives of the function.

For a function \( f(x, y) \), the necessary second partial derivatives are \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \). These derivatives help form the Hessian matrix, which is essential for using the second derivative test. The second derivative test evaluates the determinant of the Hessian matrix, denoted as \( D \), calculated by \( D = f_{xx}f_{yy} - (f_{xy})^2 \).
  • If \( D > 0 \) and \( f_{xx} > 0 \), the critical point is a local minimum.
  • If \( D > 0 \) and \( f_{xx} < 0 \), the critical point is a local maximum.
  • If \( D < 0 \), the critical point is a saddle point, indicating a change in concavity.
  • If \( D = 0 \), the test is inconclusive.
In our exercise, with \( D = -4 \), the point \( (1, 2) \) is classified as a saddle point. This test provides insight into the function's curvature and helps classify each critical point accordingly.
Saddle Point
Saddle points are unique features in the context of multivariable calculus. They occur at critical points where the surface of a function doesn't have a peak or a valley but instead behaves like a saddle.

In simpler terms, a saddle point is where the function increases in one direction and decreases in another. This results in a point that is neither a local maximum nor a minimum.

Using the second derivative test, if the determinant of the Hessian matrix \( D \) is less than zero, it indicates that the critical point is a saddle point. In our example, the determinant \( D = -4 \) confirms the critical point \( (1, 2) \) is a saddle point. This means the graph of \( f(x, y) = x^{2}-y^{2}-2x+4y+6 \) will curve up in one direction and down in another at this point. Understanding saddle points helps in modeling and analyzing surfaces in a realistic manner.
Hessian Matrix
The Hessian matrix is a square matrix comprising second-order partial derivatives of a function. It's a crucial tool in multivariable calculus for examining the behavior of functions at critical points.

To form the Hessian matrix for a two-variable function \( f(x, y) \), calculate the second derivatives: \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \). The matrix is organized as follows:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\]The Hessian matrix helps in applying the second derivative test, especially in determining a critical point's nature — whether it is a local maximum, a local minimum, or a saddle point.
  • In our exercise: \( f_{xx} = 2 \), \( f_{yy} = -2 \), and \( f_{xy} = 0 \).
With these values, we calculate the determinant \( D = 2 \, (-2) - 0^2 = -4 \), aiding in classifying the critical point as a saddle point. The Hessian matrix simplifies the process of analyzing the second derivatives by providing a structured approach to understand the function's curvature at specific points.

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