Saddle Point Analysis Of F(x, Y) At M(7, 7)

Introduction to Saddle Points and Multivariable Functions

In the realm of multivariable calculus, understanding the behavior of functions with multiple independent variables is crucial. One fascinating aspect is the identification of critical points, which are points where the function's rate of change in all directions is zero or undefined. Among these critical points, we find local maxima, local minima, and saddle points. This article delves into the specifics of determining whether a given point is a saddle point, focusing on the function f(x, y) = x³ - 147x - y² + 14y + 23 and the point M(7, 7).

Before we dive into the specifics of our problem, let’s first establish a solid understanding of saddle points. A saddle point is a critical point where the function has a minimum value along one direction and a maximum value along another direction. Visualize a saddle: it curves upwards in the direction where a rider would sit but curves downwards along the horse's spine. Mathematically, a saddle point is a point where all first-order partial derivatives are zero, but the function does not attain a local maximum or minimum. This means the point is a critical point but not an extremum.

To identify saddle points, we employ the second derivative test. This test involves calculating the second partial derivatives of the function and forming the Hessian matrix. The determinant of this matrix, evaluated at the critical point, provides crucial information about the nature of the point. If the determinant is negative, the point is a saddle point. If it’s positive, we need further analysis to determine if it’s a local maximum or minimum. If it’s zero, the test is inconclusive. The second derivative test is a powerful tool, but it relies on a good understanding of partial derivatives and the Hessian matrix. Without this foundational knowledge, applying the test can be confusing and lead to incorrect conclusions. Remember, the goal is to thoroughly analyze the function's behavior around the critical point to classify it accurately. Now, let's explore the function in question and apply these concepts to the given point M(7, 7).

Analyzing f(x, y) = x³ - 147x - y² + 14y + 23 and Identifying Critical Points

Let's consider the function f(x, y) = x³ - 147x - y² + 14y + 23. The first step in determining whether M(7, 7) is a saddle point is to find the critical points of the function. Critical points are those where the gradient of the function is zero, meaning both the first partial derivatives with respect to x and y are zero. This is because at these points, the function's rate of change is zero, indicating a potential maximum, minimum, or saddle point.

To find these points, we first compute the partial derivatives. The partial derivative with respect to x, denoted as ∂f/∂x, represents the rate of change of the function as x varies while y is held constant. Similarly, the partial derivative with respect to y, ∂f/∂y, represents the rate of change as y varies while x is constant. For our function, the partial derivatives are:

∂f/∂x = 3x² - 147 ∂f/∂y = -2y + 14

Setting these partial derivatives equal to zero gives us a system of equations:

3x² - 147 = 0 -2y + 14 = 0

Solving the second equation for y is straightforward: -2y + 14 = 0 implies y = 7. This confirms that the y-coordinate of our potential saddle point M(7, 7) satisfies this condition. Now, let's solve the first equation for x: 3x² - 147 = 0 can be simplified to x² = 49, which yields two possible solutions: x = 7 and x = -7. Therefore, we have two critical points: (7, 7) and (-7, 7). We are particularly interested in the point (7, 7) as stated in the problem.

Once we have identified the critical points, the next crucial step is to classify them. This classification will tell us whether the point is a local maximum, a local minimum, or a saddle point. To do this, we will employ the second derivative test, which involves computing the second partial derivatives and evaluating the determinant of the Hessian matrix at the critical point. Understanding these steps is essential for correctly classifying the critical points and determining the nature of the function's behavior around these points. This careful analysis will ultimately help us determine if M(7, 7) is indeed a saddle point for the given function.

Applying the Second Derivative Test

Now that we have identified the critical points, including M(7, 7), we need to apply the second derivative test to classify them. This test hinges on the second partial derivatives of the function and the determinant of the Hessian matrix. The second partial derivatives provide information about the concavity of the function, while the Hessian determinant helps us determine the nature of the critical point.

First, let's compute the second partial derivatives of f(x, y). We already have the first partial derivatives:

∂f/∂x = 3x² - 147 ∂f/∂y = -2y + 14

Now, we find the second partial derivatives:

∂²f/∂x² = ∂/∂x (3x² - 147) = 6x ∂²f/∂y² = ∂/∂y (-2y + 14) = -2 ∂²f/∂x∂y = ∂/∂y (3x² - 147) = 0 ∂²f/∂y∂x = ∂/∂x (-2y + 14) = 0

Notice that ∂²f/∂x∂y and ∂²f/∂y∂x are equal, which is expected if the second partial derivatives are continuous (Clairaut's Theorem). Next, we construct the Hessian matrix, which is a 2x2 matrix formed by the second partial derivatives:

H(x, y) = | ∂²f/∂x² ∂²f/∂x∂y | | ∂²f/∂y∂x ∂²f/∂y² |

Substituting the values, we get:

H(x, y) = | 6x 0 | | 0 -2 |

The determinant of the Hessian matrix, denoted as D(x, y), is calculated as follows:

D(x, y) = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = (6x)(-2) - (0)² = -12x

Now, we evaluate the determinant at the critical point M(7, 7):

D(7, 7) = -12(7) = -84

The sign of this determinant is crucial. If D(7, 7) is negative, then the point (7, 7) is a saddle point. If D(7, 7) is positive, we need to examine the sign of ∂²f/∂x² at (7, 7) to determine if it's a local minimum or maximum. If D(7, 7) is zero, the test is inconclusive.

Determining the Nature of M(7, 7)

Having computed the determinant of the Hessian matrix at the point M(7, 7), we are now in a position to classify this critical point. Recall that we found D(7, 7) = -84. The second derivative test dictates that if the determinant of the Hessian matrix at a critical point is negative, then that point is a saddle point. This is a direct consequence of the function exhibiting different concavity in different directions at that point – it curves upwards in one direction and downwards in another, much like the shape of a saddle.

In our case, since D(7, 7) = -84, which is indeed negative, we can definitively conclude that the point M(7, 7) is a saddle point for the function f(x, y) = x³ - 147x - y² + 14y + 23. This result aligns with the initial understanding of saddle points and the application of the second derivative test. At this point, it’s beneficial to revisit the concept of a saddle point to reinforce the understanding of what this classification implies about the function’s behavior around M(7, 7).

A saddle point, as the name suggests, resembles the shape of a saddle. Imagine a rider sitting on a saddle; the curve rises in front and behind the rider, but it curves downwards along the sides. Similarly, at a saddle point of a multivariable function, the function will increase along one axis and decrease along another. This contrasting behavior distinguishes a saddle point from local maxima and minima, where the function either decreases or increases in all directions, respectively. Understanding this geometric interpretation aids in visualizing the nature of the function around a saddle point and solidifies the analytical result obtained from the second derivative test.

Therefore, based on our calculations and the principles of the second derivative test, the point M(7, 7) is confirmed to be a saddle point of the function f(x, y) = x³ - 147x - y² + 14y + 23. This conclusion answers the question posed in the problem and demonstrates the application of multivariable calculus techniques in identifying and classifying critical points.

Conclusion

In conclusion, we embarked on a journey to determine whether the point M(7, 7) is a saddle point of the function f(x, y) = x³ - 147x - y² + 14y + 23. Through a rigorous application of the principles of multivariable calculus, we have definitively established that it is indeed a saddle point. This determination involved several key steps, starting with finding the critical points of the function by setting the first partial derivatives to zero. We then computed the second partial derivatives and formed the Hessian matrix. The crucial step was calculating the determinant of the Hessian matrix at the point M(7, 7). Since the determinant was found to be negative (-84), the second derivative test confirmed that M(7, 7) is a saddle point.

This exercise underscores the importance of understanding critical points and their classification in multivariable calculus. Saddle points, local maxima, and local minima provide valuable insights into the behavior of functions in multiple dimensions. The second derivative test is a powerful tool in this endeavor, enabling us to analytically classify these critical points. However, the test's effectiveness hinges on a solid grasp of partial derivatives, the Hessian matrix, and the interpretation of the determinant's sign.

Moreover, visualizing the nature of a saddle point – a point where the function has a minimum in one direction and a maximum in another – aids in comprehending the mathematical results. This geometric intuition enhances our understanding and appreciation of the concepts involved. The ability to identify and classify critical points has numerous applications in various fields, including optimization problems, physics, and economics, making it a fundamental skill in mathematical analysis. Therefore, the analysis performed in this article not only answers a specific question but also highlights the broader significance of these concepts in mathematics and its applications.