Problem 41 - Precalculus Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=2+\frac{1}{x}, g(x)=\frac{1}{x}$$

Step-by-Step Solution

Answer

\(f+g(x) = 2+2x^{-1}, f-g(x)\) = 2, \(f(x) \cdot g(x)\) = 2x^{-1}+x^{-2}, \(\frac{f(x)}{x} = 2x^{-1}+x^{-2}\). The domain for all these functions is \(x \neq 0\).

Step 1: Finding \(f+g\)

Add the two given functions to find \(f+g\):\n\[f(x)+g(x)=\left(2+\frac{1}{x}\right)+\left(\frac{1}{x}\right)=2+2x^{-1}\]

Step 2: Finding \(f-g\)

Subtract \(g(x)\) from \(f(x)\) to find \(f-g\):\n\[f(x)-g(x)=\left(2+\frac{1}{x}\right)-\left(\frac{1}{x}\right)=2\]

Step 3: Finding fg

Multiply the two functions to find fg:\n\[f(x)\cdot g(x)=\left(2+\frac{1}{x}\right)\cdot \left(\frac{1}{x}\right)=2x^{-1}+x^{-2}\]

Step 4: Finding \(\frac{f}{x}\)

Divide the function \(f(x)\) by \(x\) to find \(\frac{f}{x}\):\n\[\frac{f(x)}{x}=\frac{2+\frac{1}{x}}{x}=2x^{-1}+x^{-2}\]

Step 5: Finding Domains

The domain of a function is the set of all real numbers except those for which the function is undefined. None of these results are defined for \(x=0\). So, the domain for all these functions is \(x\neq0\).

Key Concepts

Function Operations

Function operations involve performing arithmetic operations on two functions, such as addition, subtraction, multiplication, and division. This allows us to create new functions from existing ones and explore their properties. Here's how these operations work:

• Addition: To add two functions, simply add their respective values. For example, to find \(f+g\) when \(f(x) = 2 + \frac{1}{x}\) and \(g(x) = \frac{1}{x}\), you combine the expressions: \(f(x) + g(x) = (2 + \frac{1}{x}) + \frac{1}{x} = 2 + 2x^{-1}\).


• Subtraction: Subtracting functions is just as straightforward. To find \(f-g\), subtract \(g(x)\) from \(f(x)\): \((2 + \frac{1}{x}) - \frac{1}{x} = 2\). This simplifies nicely to a constant.


• Multiplication: When you multiply two functions, you multiply each term. For \(fg\), multiply \(f(x)\) and \(g(x)\): \((2 + \frac{1}{x}) \cdot \frac{1}{x} = 2x^{-1} + x^{-2}\).


• Division: Dividing a function by a variable or another function involves division of each part. For example, \(\frac{f}{x} = \frac{2+\frac{1}{x}}{x} = 2x^{-1} + x^{-2}\).

Domain

The domain of a function is the set of all possible input values (usually \(x\) values) for which the function is defined. Understanding domain is essential because it tells us which values will not result in undefined expressions or mathematical errors.
For rational functions like \(f(x) = 2 + \frac{1}{x}\) and \(g(x) = \frac{1}{x}\), the domain is determined by looking for values that cause division by zero.

• If we substitute \(x = 0\) into either function, we'll end up with a term containing division by zero, which is undefined.


• Therefore, for \(f+g\), \(f-g\), \(fg\), and \(\frac{f}{x}\), the domain excludes \(x = 0\).

Thus, the domain for all these function results is the set of all real numbers except zero: \(x eq 0\). Ensuring clarity about the domain helps in graphing and more advanced function analysis.

Rational Functions

Rational functions are formed when one polynomial is divided by another. These functions frequently involve fractions where the denominator contains a variable. Recognizing them is key in modifying and analyzing various mathematical expressions.
For example, in this exercise, \(f(x) = 2 + \frac{1}{x}\) and \(g(x) = \frac{1}{x}\) are rational functions because they involve fractions where \(x\) appears in the denominator. Here are some important features of rational functions:

• Discontinuities: Rational functions can be undefined at points that make the denominator zero. In our example, both \(f(x)\) and \(g(x)\) stay undefined at \(x=0\).


• Simplification: When performing operations like addition or multiplication, simplifying expressions involves combining like terms and rationalizing any complex fractions.


• Applications: Rational functions appear in various real-world modeling situations, such as rates of change, economics, and physics.

Understanding how to manipulate and find the domain of these types of functions is crucial for solving complex equations and interpreting their graphical behaviors.