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Show that \(f(x, y, z)=x+y-z\) is continuous at every point \(\left(x_{0}, y_{0}, z_{0}\right) .\)

Short Answer

Expert verified
The function is continuous at every point \((x_0, y_0, z_0)\) in its domain.

Step by step solution

01

Understand Continuity in Multivariable Functions

To show that a function is continuous at a point, we need to demonstrate that for all \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2} < \delta \), then \( |f(x,y,z) - f(x_0,y_0,z_0)| < \epsilon \).
02

Define the Function and Points

Given the function \( f(x,y,z) = x+y-z \), we need to determine the continuity at an arbitrary point \( (x_0, y_0, z_0) \). The value of the function at this point is \( f(x_0, y_0, z_0) = x_0 + y_0 - z_0 \).
03

Compute the Difference

Calculate the difference \( |f(x, y, z) - f(x_0, y_0, z_0)| = |(x+y-z) - (x_0 + y_0 - z_0)| = |(x - x_0) + (y - y_0) - (z - z_0)| \).
04

Apply the Triangle Inequality

Using the triangle inequality, we find that \( |(x - x_0) + (y - y_0) - (z - z_0)| \leq |x - x_0| + |y - y_0| + |z - z_0| \).
05

Define the Relation Between \(\delta\) and \(\epsilon\)

Choose \( \delta = \frac{\epsilon}{3} \). Then if \( \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2} < \delta \), we have \( |x - x_0| < \delta \), \( |y - y_0| < \delta \), and \( |z - z_0| < \delta \).
06

Establish the Final Inequality

It follows that \( |x - x_0| + |y - y_0| + |z - z_0| < 3\delta = \epsilon \). Therefore, \( |f(x, y, z) - f(x_0, y_0, z_0)| < \epsilon \), proving continuity.

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Key Concepts

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

Epsilon-Delta Definition
The epsilon-delta definition of continuity is pivotal in understanding how to demonstrate that a function is continuous at a given point. Let's break this down to lay the groundwork for clarity.

In the realm of single-variable calculus, a function is said to be continuous at a point if the output value of the function approaches the expected limit as the input approaches that specific point. When we transition to multivariable calculus, this idea needs slight refinement. Now, we consider not one, but multiple inputs.

The epsilon-delta definition provides the rigorous foundation for this concept. For any multivariable function, say, \( f(x, y, z) \), we're tasked to show that for every small error range, or \( \epsilon > 0 \), there exists a small distance, or \( \delta > 0 \), such that whenever the input \( (x, y, z) \) is within \( \delta \) of the point \( (x_0, y_0, z_0) \), the difference between the output of the function \( f(x, y, z) \) and \( f(x_0, y_0, z_0) \) remains within the error range \( \epsilon \). In other words:
  • If \( \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2} < \delta \)
  • Then \( |f(x, y, z) - f(x_0, y_0, z_0)| < \epsilon \)
By establishing such a delta for all potential epsilons, the function \( f(x, y, z) = x+y-z \) is shown to be continuous at every point \( (x_0, y_0, z_0) \). Each calculation becomes an exercise in ensuring the function behaves calmly and predictably as it approaches any given point.
Triangle Inequality
The triangle inequality is a useful tool in assessing how differences in inputs of a function can affect its output. Specifically, it’s a way to measure how adding components (or distances) behaves.

In the context of proving continuity, the triangle inequality helps us estimate the upper limits of potential deviation. Let’s delve into how this is applied to our specific scenario.

For our function \( f(x, y, z) = x+y-z \), we need to prove that the difference in output, or \( |f(x, y, z) - f(x_0, y_0, z_0)| \), is limited. By calculating the difference as \( |(x-x_0) + (y-y_0) - (z-z_0)| \), we recognize that this expression reflects the total deviation in the function's outcome arising from slight shifts in the inputs \( x, y, \text{and} z \).

The triangle inequality helps us by allowing:
  • \( |(x-x_0) + (y-y_0) - (z-z_0)| \leq |x-x_0| + |y-y_0| + |z-z_0| \)
This makes it easier to further control the function's potential fluctuation within the specified limits of \( \epsilon \). This understanding is vital in ensuring the function behaves "nicely" near every point \((x_0, y_0, z_0)\), contributing to its continuity proof.
Multivariable Calculus
Multivariable calculus extends the ideas of single-variable calculus to functions with multiple inputs. This branch of mathematics allows us to explore a richer array of problems by considering functions that take several variables as inputs, such as \( f(x, y, z) = x+y-z \).

Continuity in this domain becomes more intricate due to the involvement of multiple dimensions. A multivariable function's continuity at a point means that small changes in its multiple variables result in a correspondingly small change in the function's value.

Understanding continuity in multivariable contexts requires visualizing changes in multi-dimensional space. Instead of a single input-variable line approaching a point, we watch how a point in 3D space (like a point within \((x, y, z)\)) closes in on another fixed point \((x_0, y_0, z_0)\). Hence, the mathematical conditions become those defined by the epsilon-delta framework. This assures that as the inputs come together towards the target point, the function’s output stays predictably close to its actual value at that point. Multivariable calculus, therefore, provides the essential tools and concepts necessary for understanding real-world scenarios where multiple factors affect an outcome, equipping learners to handle complex systems.

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Most popular questions from this chapter

Box with vertex on a plane Find the volume of the largest closed rectangular box in the first octant having three faces in the coordinate planes and a vertex on the plane \(x / a+y / b+z / c=1\) where \(a>0, b>0,\) and \(c>0\).

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I,\) you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$ \begin{array}{l}{x=2 \cos u \cos v, \quad y=2 \cos u \sin v, \quad z=2 \sin u} \\\ {0 \leq u \leq 2 \pi, \quad 0 \leq v \leq \pi}\end{array} $$

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z, w)=x^{2}+y^{2}+z^{2}+w^{2}\) subject to the constraints \(\quad 2 x-y+z-w-1=0 \quad\) and \(\quad x+y-z+\) \(w-1=0.\)

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I,\) you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$ \begin{array}{l}{x=u \cos v, \quad y=u \sin v, \quad z=u, \quad 0 \leq u \leq 2} \\ {0 \leq v \leq 2 \pi}\end{array} $$

Let $$f(x, y)=\left\\{\begin{array}{ll}{0,} & {x^{2} < y<2 x^{2}} \\ {1,} & {\text { otherwise }}\end{array}\right.$$ Show that \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist, but \(f\) is not differentiable at \((0,0) .\)

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