Problem 1 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The assumption that \((1,1)\) is a critical point contradicts \(f_y(1,1) \neq 0\); unresolved processes suggest data need reevaluation or error indication.

Step 1: Identify the Information Given

For (a), we know \( f_{xx}(1,1) = 4 \), \( f_{xy}(1,1) = 1 \), and \( f_{y}(1,1) = 2 \). For (b), we have \( f_{xx}(1,1) = 4 \), \( f_{xy}(1,1) = 3 \), and \( f_{y}(1,1) = 2 \).

Step 2: Verify Condition for Critical Points

A critical point, \((1,1)\), means both first partial derivatives must be zero at this point. Check data: \( f_y(1,1) \) should be zero for a point to be critical in this context. Here, \( f_y(1,1) = 2 \) indicates it cannot be a critical point based on \( f_y \).

Step 3: Consider the Mistaken Critical Point Indication

Since \( f_y(1,1) = 2 \), actually \((1, 1)\) is not truly a critical point unless \( f_x(1,1) \) is revealed to be zero, possibly suggesting merely an error in citing or a misalignment between problem statement and assumed calculation.

Key Concepts

Second Partial Derivatives

Second partial derivatives are crucial when analyzing critical points in functions of two variables. These derivatives, denoted by \( f_{xx} \), \( f_{xy} \), and \( f_{yy} \), provide information about the concavity of a function's graph at a given point.

The second partial derivative \( f_{xx} \) indicates how the function changes as the x-variable changes. Similarly, \( f_{yy} \) shows changes relative to the y-variable. The mixed partial derivative, \( f_{xy} \), reveals how the function changes in response to both variables interacting.

• If the derivatives are all positive or all negative, the point may be a minimum or maximum.
• Contrasting signs can indicate a saddle point, where the function rises in one direction and falls in another.

Understanding these interactions is essential for determining the nature of a critical point. If \( f_y(1,1) \) cannot be zero, the original assumption about the critical point might have errors affecting subsequent analysis.

Function Analysis

Function analysis involves examining key features of a function, such as continuity and differentiability, to understand its behavior and critical points. A function with continuous second derivatives is differentiable in a smooth, predictable manner.

When assessing critical points like \(1, 1\) for a given function \( f(x, y) \), it's essential to note:

• The first partial derivatives must both be zero for a true critical point.
• The second partial derivatives, evaluated using the Hessian matrix, classify the critical point further.

Calculations should align with the problem's characteristics and given data. Errors like assuming \( f_y(1,1) = 2 \) indicate miscalculations or data misinterpretations, suggesting the point may not be critical as defined initially.

Mathematical Problem Solving

Mathematical problem solving in calculus involves a systematic approach to finding solutions, verifying conditions, and checking computations for consistency with problem requirements.

Steps to approach problems like our given scenario include:

• Clearly identify all given information and assumptions about a point being critical.
• Apply the first and second derivative tests to validate or refute the critical nature of the point.
• Double-check all calculations to avoid errors and ensure conclusions match initial conditions stated in the problem.

Effective problem solving requires meticulous attention to detail and a deep understanding of the mathematical principles governing the problem at hand. This ensures accurate identification of critical points and the proper application of derivative tests.