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A differentiable function \(f(x, y)\) has a saddle point at a point \((a, b)\) where its partial derivatives are simultaneously zero, if in every open disk centered at \((a, b)\) there are domain points where \(f(x, y)>f(a, b)\) and domain points where \(f(x, y)

Short Answer

Expert verified
Saddle points: (a) (0,0); (b) (1,-1).

Step by step solution

01

Find the First Partial Derivatives

To find the saddle points, first calculate the partial derivatives of the function with respect to both variables. **For function (a)**, the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 3x^2 - 2y\) and with respect to \(y\) is \(\frac{\partial f}{\partial y} = -3y^2 - 2x\). **For function (b)**, the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 12x - 6x^2 + 6y\) and with respect to \(y\) is \(\frac{\partial f}{\partial y} = 6y + 6x\).
02

Solve for Critical Points

Set each partial derivative equal to zero and solve the system of equations to find critical points. **For function (a):** Solve \(3x^2 - 2y = 0\) and \(-3y^2 - 2x = 0\). **For function (b):** Solve \(12x - 6x^2 + 6y = 0\) and \(6y + 6x = 0\).
03

Critical Points for Function (a)

For \(3x^2 = 2y\) and \(-3y^2 = 2x\), substitute \(y = \frac{3}{2}x^2\) into the second equation. This results in \(-\frac{27}{4}x^4 = 2x\). Solving, \(x(x^3\frac{27}{4} + 2) = 0\) gives \(x=0\) or \(x^3\frac{27}{4} + 2 = 0\). Substitute \(x=0\) into the equation for \(y\), yielding \(y = 0\), thus the critical point is \((0,0)\).
04

Critical Points for Function (b)

For \( -6x^2 + 12x + 6y = 0\) and \(6x + 6y = 0\), from the second equation, substitute \(y = -x\) into the first: \(-6x^2 + 12x - 6x = 0\) simplifies to \(-6x^2 + 6x = 0\), or \(6x(x-1) = 0\). Thus, \(x = 0\) or \(x = 1\). So, critical points are \((0, 0)\) and \((1, -1)\).
05

Determine Saddle Points with the Second Derivative Test

For function (a), calculate second derivatives \(f_{xx}=6x\), \(f_{yy}=-6y\), and \(f_{xy}=-2\). Second derivative test: \(D = f_{xx}f_{yy} - (f_{xy})^2 = (6x)(-6y)-(-2)^2\). At \((0,0)\), \(D = 0\) which is inconclusive, but check the definition: \(f(x,y)\) changes sign around \((0,0)\). Thus, \((0, 0)\) is a saddle point. For function (b), second derivatives are \(f_{xx}=-12x+12\), \(f_{yy}=6\), and \(f_{xy}=6\). Second derivative test for \((0,0): D = (12)(6)-(6)^2=36 > 0\), non-saddle. \((1,-1): D = (-12 + 12)(6) - (6)^2 = -36\), a saddle point.

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Key Concepts

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

Differentiable Function
A differentiable function is a smooth, continuous function that has a derivative at every point in its domain. This essentially means that the function has no sharp corners or abrupt changes in direction. In mathematical terms, if a function is differentiable at a point, the rate of change of the function at that point is well-defined. Differentiability is a key concept in calculus because it allows us to use derivatives to analyze and predict the behavior of functions.
In the context of multi-variable functions like those in our saddle point problem, differentiability implies that we can take partial derivatives with respect to each variable. For instance, for a function \(f(x, y)\), differentiability allows us to compute \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). Understanding how a function changes in response to small changes in \(x\) and \(y\) is crucial to finding critical points and identifying features like saddle points.
Partial Derivatives
Partial derivatives are derivatives of functions of multiple variables with respect to one variable while holding the other variables constant. This is crucial when analyzing how a function behaves in different directions at any given point.
  • For instance, consider the function \(f(x, y)\). The partial derivative of \(f\) with respect to \(x\) is given by \(\frac{\partial f}{\partial x}\), which represents how \(f\) changes as \(x\) changes, while \(y\) is kept constant.
  • Similarly, \(\frac{\partial f}{\partial y}\) tells us how \(f\) changes as \(y\) changes, with \(x\) held constant.
Calculating partial derivatives is a crucial step in identifying critical points, which are candidates for local maxima, minima, or saddle points.
In our exercise, we first found the partial derivatives of each function and solved them to zero to find the critical points. This method helps in detecting where the function doesn't have a clear increasing or decreasing tendency, a potential indicator of a saddle point.
Second Derivative Test
The second derivative test is a method to classify critical points of differentiable functions of several variables. It helps us determine whether a given critical point is a local maximum, local minimum, or saddle point.
  • First, calculate the second partial derivatives: \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\). These represent how the rates of change in one direction change as you move in other directions.
  • The discriminant, denoted as \(D\), is computed as \(D = f_{xx}f_{yy} - (f_{xy})^2\).
If \(D > 0\), and \(f_{xx} > 0\), the point is a local minimum. If \(D > 0\), and \(f_{xx} < 0\), it's a local maximum. But if \(D < 0\), the point is a saddle point. When \(D = 0\), the test is inconclusive. This test was applied in our exercise to distinguish between saddle points and other critical points for the given functions.

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