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Chapter 9: Problem 1

Suppose \((1,1)\) is a critical point of a function \(f\) with continuous second derivatives. In each case, what can you say about \(f ?\) \(\begin{array}{ll}{\text { (a) } f_{x x}(1,1)=4,} & {f_{x y}(1,1)=1, \quad f_{y y}(1,1)=2} \\ { (b) f_{x x}(1,1)=4,} & {f_{x y}(1,1)=3, \quad f_{y y}(1,1)=2}\end{array}\)

Short Answer

Expert verified
In part (a), \((1,1)\) is a local minimum; in part (b), it is a saddle point.

Step by step solution

01

Understand Critical Points

A critical point occurs where the first partial derivatives of a function are zero. At the point \((1,1)\), the first derivatives \(f_x(1,1)\) and \(f_y(1,1)\) must be zero for it to be a critical point.
02

Hessian Determinant for Classification

To determine the nature of a critical point, we examine the Hessian determinant, given by \( D = f_{xx}f_{yy} - (f_{xy})^2 \). A positive \(D\) implies the point is a local minima or maxima, while a negative \(D\) suggests a saddle point.
03

Calculate Part (a)

For part (a), we have: \(f_{xx}(1,1) = 4\), \(f_{xy}(1,1) = 1\), \(f_{yy}(1,1) = 2\). Therefore, the Hessian determinant is \(D = 4 \times 2 - 1^2 = 8 - 1 = 7\). Since \(D > 0\) and \(f_{xx} > 0\), the point \((1,1)\) is a local minimum.
04

Calculate Part (b)

For part (b), we have: \(f_{xx}(1,1) = 4\), \(f_{xy}(1,1) = 3\), \(f_{yy}(1,1) = 2\). Therefore, the Hessian determinant is \(D = 4 \times 2 - 3^2 = 8 - 9 = -1\). Because \(D < 0\), the point \((1,1)\) is a saddle point.

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Key Concepts

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

Hessian Determinant
To analyze the nature of critical points, we often use the Hessian determinant. The Hessian determinant helps determine whether a critical point is a local extremum or a saddle point. The Hessian is a matrix composed of all the second partial derivatives of a function. For a function of two variables, say \(f(x, y)\), the Hessian matrix \(H\) is structured as follows:
  • \(H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\)
From this, the Hessian determinant \(D\) is calculated using the formula:
  • \(D = f_{xx}f_{yy} - (f_{xy})^2\)
The determinant helps classify the critical point:
  • If \(D > 0\) and \(f_{xx} > 0\), the point is a local minimum.
  • If \(D > 0\) and \(f_{xx} < 0\), the point is a local maximum.
  • If \(D < 0\), the point is a saddle point.
  • If \(D = 0\), the test is inconclusive.
This process helps in understanding the behavior of the function around critical points.
Second Derivatives
Second derivatives play a crucial role in identifying the nature of a critical point through the Hessian matrix. A second derivative measures how the rate of change of a function's slope is itself changing. For multivariable functions, these second derivatives provide valuable insights:
  • \(f_{xx}\) and \(f_{yy}\) are the second partial derivatives with respect to \(x\) and \(y\), respectively. They indicate the concavity of the function in each direction.
  • \(f_{xy}\) and \(f_{yx}\), also known as mixed derivatives, describe how one variable affects the slope concerning the other variable.
These derivatives together form the Hessian matrix, aiding in the classification of the critical point. For example:
  • If \(f_{xx} > 0\), the function is concave upwards in the \(x\)-direction, suggesting a potential minimum.
  • Conversely, if \(f_{xx} < 0\), the function is concave downwards, indicating a possible maximum.
Thus, understanding second derivatives is key to evaluating the concavity and behavior of multivariable functions around critical points.
Local Extremum
A local extremum refers to points on a function where the function reaches a local maximum or minimum. At these points, the function's slope is zero and the change of slope is characterized using the second derivatives:
  • A **local maximum** occurs where the function's value is greater than at any nearby point. Here, the slope transitions from positive to negative.
  • A **local minimum**, conversely, occurs where the function's value is less than at any nearby points, with the slope shifting from negative to positive.
To determine if a critical point is a local extremum, we rely on the sign of the Hessian determinant:
  • If \(D > 0\) and since the second derivative \(f_{xx} > 0\), the critical point is a local minimum.
  • If \(D > 0\) and \(f_{xx} < 0\), it indicates a local maximum.
The local extremum gives insight into the function's behavior, indicating where it potentially peaks or dips in its local region.
Saddle Point
Saddle points are unique critical points where the function does not exhibit a local minimum or maximum. Instead, a saddle point appears as a point on a surface that acts like a minimum in one direction and a maximum in another. This behavior is distinct from typical peaks or troughs found in local extrema.
  • A typical example is a Pringles chip-shaped surface, where at the center, pressing down pushes one part of the chip upwards and the other downwards.
A saddle point can be identified using the Hessian determinant:
  • If the Hessian determinant \(D < 0\), it signals a saddle point, as seen in problem part (b), where the calculation resulted in \(D = -1\).
The determination of a saddle point is crucial as it represents points where the function does not definitively rise or fall in value relative to its surrounding points, making them critical in optimization and further mathematical analysis.

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