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If \(f_{x}(a, b)=f_{y}(a, b)=0,\) must \(f\) have a local maximum or minimum value at \((a, b) ?\) Give reasons for your answer.

Short Answer

Expert verified
No, it could be a saddle point if the second derivative test shows \(D < 0\).

Step by step solution

01

Understand Critical Points

A point \((a, b)\) is considered a critical point of a function \(f(x, y)\) if the first partial derivatives \(f_x(a, b)\) and \(f_y(a, b)\) are both zero. Critical points are potential candidates for local maxima, minima, or saddle points.
02

The Second Derivative Test

To determine the nature of the critical point \((a, b)\), use the second derivative test. Compute the second partial derivatives: \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\) at \((a, b)\).
03

Compute the Discriminant

Calculate the discriminant \(D\) using the formula \(D = f_{xx}(a, b) \cdot f_{yy}(a, b) - [f_{xy}(a, b)]^2\). The value of \(D\) will help determine the critical point's classification.
04

Analyze the Discriminant

- If \(D > 0\) and \(f_{xx} > 0\), \((a, b)\) is a local minimum. - If \(D > 0\) and \(f_{xx} < 0\), \((a, b)\) is a local maximum. - If \(D < 0\), \((a, b)\) is a saddle point, not a maximum or minimum. - If \(D = 0\), the test is inconclusive, so further analysis is required.

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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 like regular derivatives, but they apply to functions with more than one variable. For a function of two variables, say \(f(x, y)\), the partial derivative with respect to \(x\) (denoted as \(f_x\)) is found by holding \(y\) constant and differentiating with respect to \(x\). Similarly, the partial derivative with respect to \(y\) (denoted as \(f_y\)) is found by holding \(x\) constant and differentiating with respect to \(y\).
Such derivatives are useful in determining how the function changes at a given point along each variable’s direction. In the context of critical points, partial derivatives are used to identify where the slope of the surface is flat, which could potentially indicate a local maximum, minimum, or a saddle point.
At a critical point, both \(f_x\) and \(f_y\) are equal to zero, suggesting no local slope and setting the stage for further investigation with the second derivative test.
Second Derivative Test
The second derivative test helps us to determine the nature of a critical point by examining the concavity of the function. This involves using second-order partial derivatives: \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\).
The second derivative \(f_{xx}\) is the partial derivative of \(f_x\) with respect to \(x\), \(f_{yy}\) is the partial derivative of \(f_y\) with respect to \(y\), and \(f_{xy}\) or \(f_{yx}\) is the mixed partial derivative, which in most cases are equal due to Clairaut's theorem.
  • If both \(f_{xx}\) and \(f_{yy}\) are positive, the function is concave up at the critical point, hinting at a local minimum.
  • If both are negative, the function is concave down, indicating a local maximum.
The test involves computing these second partial derivatives at the critical point to further proceed with understanding its nature.
Discriminant in Calculus
The discriminant in calculus involves using a specific formula that incorporates the second partial derivatives to classify critical points. The formula for the discriminant \(D\) is:
\[ D = f_{xx}(a, b) \cdot f_{yy}(a, b) - [f_{xy}(a, b)]^2 \]
This discriminant helps in determining the nature of the critical point \((a, b)\):
  • If \(D > 0\) and \(f_{xx} > 0\), we have a local minimum.
  • If \(D > 0\) and \(f_{xx} < 0\), we have a local maximum.
  • If \(D < 0\), the point is a saddle point.
  • If \(D = 0\), the test is inconclusive, requiring further investigation.

This process is crucial for determining the type of extremum or feature the function has at the critical point.
Saddle Points
Saddle points occur at critical points where \(D < 0\), suggesting neither a local maximum nor a minimum. The term comes from the way a horse saddle curves up in one direction and down in another. Saddle points often indicate a change in the nature of a function at a critical point.
They illustrate cases where the function curves upwards in one direction, and simultaneously curves downwards in an orthogonal direction. This characteristic ensures the point is genuinely flat (as both first partial derivatives are zero), yet it is not an extremum.
Recognizing a saddle point is crucial in multivariable calculus and critical in optimization problems, as it can often be misleading if one only looks at the fact that the partial derivatives are zero. Therefore, understanding additional factors, such as the discriminant, is important in distinguishing these critical points.

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Most popular questions from this chapter

Suppose that the equation \(g(x, y, z)=0\) determines \(z\) as a differentiable function of the independent variables \(x\) and \(y\) and that \(g_{z} \neq 0 .\) Show that $$ \left(\frac{\partial z}{\partial y}\right)_{x}=-\frac{\partial g / \partial y}{\partial g / \partial z} $$

Find the absolute maximum and minimum values of the following functions on the given curves. Functions: $$ \begin{array}{ll}{\text { a. }} & {f(x, y)=2 x+3 y} \\ {\text { b. }} & {g(x, y)=x y} \\ {\text { c. }} & {h(x, y)=x^{2}+3 y^{2}}\end{array} $$ Curves: i) The semiellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad y \geq 0\) ii) The quarter ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=3 \cos t, y=2 \sin t\)

Find the maximum value of \(s=x y+y z+x z\) where \(x+y+z=6 .\)

In Exercises \(71-76\) , you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$ \begin{array}{l}{f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3}} , {-4 \leq x \leq 3,-2 \leq y \leq 2}\end{array} $$

The fifth-order partial derivative \(\partial^{5} f / \partial x^{2} \partial y^{3}\) is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: \(x\) or \(y ?\) Try to answer without writing anything down. \begin{equation}\begin{array}{ll}{\text { a. }} & {f(x, y)=y^{2} x^{4} e^{x}+2} \\ {\text { b. }} & {f(x, y)=y^{2}+y\left(\sin x-x^{4}\right)} \\\ {\text { c. }} & {f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { d. }} & {f(x, y)=x e^{y^{2} / 2}}\end{array}\end{equation}

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