Problem 39 - Calculus Early Transcendentals

Question

Show that the Second Derivative Test is inconclusive when applied to the following functions at \((0,0) .\) Describe the behavior of the function at the critical point. $$f(x, y)=4+x^{4}+3 y^{4}$$

Step-by-Step Solution

Answer

Answer: Yes, the Second Derivative Test is inconclusive for the critical point (0,0) as the discriminant, D, equals 0.

Step 1: Find the first partial derivatives

We need to find the first-order partial derivatives with respect to \(x\) and \(y\): $$f_x = \frac{\partial f}{\partial x} = 4x^3$$ and $$f_y = \frac{\partial f}{\partial y} = 12y^3$$

Step 2: Identify the critical points

Critical points are where both \(f_x\) and \(f_y\) are equal to zero, or undefined. In this case, both partial derivatives are defined for all values of \(x\) and \(y\), so we set them to zero and solve for \(x\) and \(y\): $$4x^3 = 0 \Rightarrow x = 0$$ $$12y^3 = 0 \Rightarrow y = 0$$ Thus, we have a critical point at \((0,0)\).

Step 3: Compute the second partial derivatives

Now, we will compute the second-order partial derivatives: $$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x^2$$ $$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 36y^2$$ $$f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0$$ $$f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = 0$$

Step 4: Apply the Second Derivative Test

The Second Derivative Test uses the discriminant, defined as: $$D = (f_{xx} \cdot f_{yy} ) - (f_{xy} \cdot f_{yx} )$$ At the critical point \((0,0)\), we have: $$D =(12 \cdot 0^2)(36 \cdot 0^2) - (0)(0) = 0$$ Since \(D=0\), the Second Derivative Test is inconclusive. That means we cannot determine if the critical point is a maximum, minimum, or saddle point by simply analysing the second partial derivatives.

Step 5: Describe the behavior of the function at the critical point

Although the Second Derivative Test is inconclusive, we can still get a sense of the behavior of \(f(x,y)\) near the critical point by examining the general form of the function: $$f(x,y) = 4 + x^4 + 3y^4$$ Observe that for any \(x\) and \(y\), \(x^4 \ge 0\) and \(3y^4 \ge 0\), so the sum of these terms is always nonnegative. Additionally, the presence of the \(x^4\) and \(y^4\) terms implies that the function grows indefinitely as \(x\) and \(y\) move away from the critical point. Since the function exhibits no sign changes (it is always positive), the function is neither a maximum nor a saddle point at the critical point, but rather a local minimum. Note that this conclusion is not derived from the Second Derivative Test but from an analysis of the function itself.

Key Concepts

Partial Derivatives

To understand the Second Derivative Test, we first need to delve into the concept of partial derivatives. When dealing with a function of multiple variables, like our function \(f(x, y) = 4 + x^4 + 3y^4\), partial derivatives help us understand how the function changes as we vary just one of those variables. In this case, the partial derivative with respect to \(x\) tells us how the function \(f\) changes when only \(x\) changes, keeping \(y\) constant. Similarly, the partial derivative with respect to \(y\) shows the change in \(f\) when \(y\) changes, keeping \(x\) constant.

For our function:

• The partial derivative concerning \(x\), denoted \(f_x\), is \(f_x = 4x^3\).
• The partial derivative concerning \(y\), denoted \(f_y\), is \(f_y = 12y^3\).

These expressions help in assessing changes in different directions independently, which is crucial for locating points of interest in multivariable calculus, like critical points.

Critical Points

Critical points are where the first partial derivatives of a function are either zero or undefined. These points are significant because they can indicate potential local maxima, minima, or saddle points for the function.

In the case of \(f(x, y) = 4 + x^4 + 3y^4\), we found the critical point by setting both partial derivatives \(f_x = 0\) and \(f_y = 0\). Solving these equations gives \(x = 0\) and \(y = 0\), leading us to a critical point at \((0, 0)\).

Understanding these points allows us to further analyze the behavior of the function around these locations, forming the foundation for more advanced tests, like the Second Derivative Test, which can provide additional insights into the nature of these points.

Discriminant

In the Second Derivative Test, the discriminant plays a critical role in assessing the nature of critical points. The discriminant, in this context, is a value derived from second-order partial derivatives and is crucial for determining the character of critical points

The formula used is:

\[D = f_{xx} \, f_{yy} - (f_{xy})^2\]

For our function at the critical point \((0,0)\), the second-order partial derivatives are:

• \(f_{xx} = 12x^2\)
• \(f_{yy} = 36y^2\)
• \(f_{xy} = f_{yx} = 0\)

Plugging these into the discriminant formula gives us \(D = 0\), which means the Second Derivative Test is inconclusive. This occurs because a discriminant of zero leaves the classification of the critical point ambiguous. In such cases, alternative methods or further analysis of the function itself may be necessary to understand the behavior at that point.