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Find all critical points of the following functions. \(f(x, y)=1+x^{2}+y^{2}\)

Short Answer

Expert verified
Answer: The function has a local minimum at the point (0, 0).

Step by step solution

01

Compute the partial derivatives

We need to first find the partial derivatives of the function with respect to both x and y. Let \(f(x, y) = 1 + x^2 + y^2\). The partial derivatives are: $$ \begin{aligned} \frac{\partial f}{\partial x} = 2x, \\ \frac{\partial f}{\partial y} = 2y. \end{aligned} $$
02

Find critical points

To find the critical points, we need to find the points where both partial derivatives are equal to zero or undefined. In this case, the partial derivatives are always defined: $$ \begin{aligned} 2x = 0, \\ 2y = 0. \end{aligned} $$ Solving these equations gives us \(x = 0\) and \(y = 0\). Thus, the only critical point is \((0, 0)\).
03

Classify the critical points

Now, we will use the second partial derivative test to classify the critical point. We need to compute the following second partial derivatives: $$ \begin{aligned} \frac{\partial^2 f}{\partial x^2} = 2, \\ \frac{\partial^2 f}{\partial x \partial y} = 0, \\ \frac{\partial^2 f}{\partial y^2} = 2. \end{aligned} $$ Now, calculate the discriminant: $$ D = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2. $$ Plugging in the second partial derivatives, we get: $$ D = (2)(2) - (0)^2, $$ which simplifies to \(D = 4\). Since \(D > 0\) and \(\frac{\partial^2 f}{\partial x^2} > 0\), the critical point \((0, 0)\) is a local minimum. Thus, the function \(f(x, y) = 1 + x^2 + y^2\) has a local minimum at the point \((0, 0)\).

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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 a crucial tool in multivariable calculus, representing how a function changes with respect to one of the variables, keeping the others constant. In other words, they show the slope of the function along the axis of one variable.

For instance, consider the function, \( f(x, y) = 1 + x^2 + y^2 \). To find the rate at which \( f \) changes in the direction of \( x \), we compute the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} = 2x \). Similarly, we derive \( \frac{\partial f}{\partial y} = 2y \) to represent the change along the \( y \) direction.

In identifying critical points, the locations where these derivatives are zero become of particular interest, as they might indicate peaks, valleys or saddle points on the surface defined by the function.
Second Partial Derivative Test
Once we've located critical points by setting the partial derivatives to zero, we need a reliable method to classify them. This is where the second partial derivative test comes into play. It involves taking the second derivatives of the function.

In our example, we have second derivatives, \( \frac{\partial^2 f}{\partial x^2} = 2 \), \( \frac{\partial^2 f}{\partial x \partial y} = 0 \), and \( \frac{\partial^2 f}{\partial y^2} = 2 \). These values tell us about the concavity of the function in the directions of \( x \) and \( y \), as well as any twisting of the surface.

The test can confirm whether a critical point is a local maximum, a local minimum, or a saddle point. This is crucial for understanding the topography of the function's graph and for predicting how the function behaves around these points.
Discriminant in Multivariable Calculus
The discriminant in multivariable calculus is a value computed from the second partial derivatives, which helps in using the second partial derivative test effectively. For a function with two variables, the discriminant is given by \( D = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \).

A positive discriminant combined with a positive second derivative with respect to \( x \) indicates a local minimum, while a negative discriminant suggests a saddle point. If \( D \) is positive and the second derivative with respect to \( x \) is negative, it's a local maximum. In our function's case, we calculated \( D = 4 \), which supports the conclusion of a local minimum at the critical point.
Local Minimum
When we speak of a local minimum in the context of multivariable functions, we refer to a point where the function value is lower than that of any other nearby points.

The second partial derivative test, along with the discriminant, can confirm such a point, given certain conditions are met. In our example, the function \( f(x, y) \) was determined to have a local minimum at the critical point \( (0, 0) \), suggesting a small 'valley' on the function's surface around this point. Understanding local minima is crucial for applications involving optimization, as they can represent the most efficient, least costly, or least risky solution to a problem within a specific region.

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