Problem 11 - The Calculus with Analytic Geometry

Question

In Exercises 3 through 11 , find the domain and range of the function \(f\) and draw a sketch showing as a shaded region in \(R^{2}\) the set of points in the domain of \(f\). $$ f(x, y)=\ln (x y-1) $$

Step-by-Step Solution

Answer

Domain: \((x,y)\) such that \(xy > 1\). Range: \((-\infty, \infty)\).

Step 1 - Understanding the function

The given function is: \(f(x, y) = \ln(xy - 1)\). The natural logarithm function, \(\ln(z)\), is defined only for positive values of \(z\). Therefore, we need the argument inside the logarithm to be positive.

Step 2 - Set the inequality for the domain

For \(\ln(xy - 1)\) to be defined, the expression \(xy - 1\) must be greater than 0. Set up the inequality: \(xy - 1 > 0\).

Step 3 - Solve the inequality

Solve the inequality: \(xy - 1 > 0\). This simplifies to \(xy > 1\). Therefore, the domain of the function \(f(x, y)\) is the set of all \((x, y)\) such that \(xy > 1\).

Step 4 - Identify the range

Since the function is \(\ln(xy - 1)\), the range of the natural logarithm is \((-\infty, \infty)\), because the argument \(xy - 1\) can take on any value in \((0, \infty)\).

Step 5 - Sketch the domain

To sketch the domain on the \(x-y\) plane, we need to shade the region where \(xy > 1\). The boundary is the hyperbola \(xy = 1\), so shade the region above this hyperbola. The region should not include the hyperbola itself.

Step 6 - Combine the results

The domain of \(f(x, y) = \ln(xy - 1)\) is all \((x, y)\) such that \(xy > 1\). The range of this function is \((-\infty, \infty)\). The sketch will show the region where \(xy > 1\) as a shaded region above the hyperbola \(xy = 1\).

Key Concepts

Natural logarithm

The natural logarithm, \(\text{ln}(z)\), is a critical concept in mathematics, representing the logarithm with base \(e (approximately\ 2.718)\). The function is only defined for positive values of \(z\). This has significant implications for solving problems involving natural logarithms.
In the given exercise, the function \(f(x, y) = \ln(xy - 1)\) requires that \(xy - 1 > 0\). This condition ensures the expression inside the logarithm is positive.
Always remember:

• \(\text{ln}(z)\) is undefined for \(z \leq 0\)
• The domain of \(\text{ln}(z)\) is \((0, \infty)\)
• The range of \(\text{ln}(z)\) is \((-\infty, \infty)\)

Understanding these properties will make it easier to work with natural logarithms in various contexts.

Inequality solving

Inequalities are fundamental in defining domains for functions. In this exercise, we encounter the inequality \(xy - 1 > 0\). Solving this inequality will help us determine where the function \(f(x, y) = \ln(xy - 1)\) is defined. Let's break it down:

• Start with \(xy - 1 > 0\)

The first step is to isolate the product \(xy\):
\[xy - 1 > 0\]
Add 1 to both sides to get:
\[xy > 1\]
This tells us that the function is defined for all \(x, y\) pairs where \(xy\) is greater than 1.
When graphing this, it’s important to:

• Shade the region where \(xy > 1\)
• Consider the boundary, which is given by the equation \(xy = 1\)

Solve inequalities methodically to find where your functions work.

Hyperbola

A hyperbola is a type of conic section that appears frequently in various mathematical contexts, including this exercise.
To better understand the domain of the function \(f(x, y) = \ln(xy - 1)\), we need to interpret the inequality \(xy > 1\) in terms of graphing.
Here's how hyperbolas come into play:

• The equation \(xy = 1\) represents a hyperbola in the \(xy\)-plane

To sketch the domain of this function, shade the region where \(xy > 1\)
Notice that the equation \(xy = 1\) forms the boundary.
Key points about hyperbolas:

• Hyperbolas are symmetrical about the axes
• They have two branches, extending in opposite directions
• The given hyperbola \(xy = 1\) is centered at the origin (0,0)

In this exercise, the hyperbola helps us clearly visualize the area of the domain for \(\text{ln}(xy - 1)\).