Problem 6 - Algebra and Trigonometry

Question

\(1-6\) Find \(f+g, f-g, f g,\) and \(f / g\) and their domains. $$ f(x)=\frac{2}{x+1}, \quad g(x)=\frac{x}{x+1} $$

Step-by-Step Solution

Answer

Addition and subtraction have a domain \(x \neq -1\). Multiplication is undefined at \(x = -1\). Division is undefined at \(x = -1\) and \(x = 0\).

Step 1: Find f+g

To sum the functions, we write: \[f(x) + g(x) = \frac{2}{x+1} + \frac{x}{x+1}\] Since they have the same denominator, we can simply add the numerators:\[f(x) + g(x) = \frac{2 + x}{x+1}\] The domain excludes \(x = -1\) due to division by zero.

Step 2: Find f-g

To find the difference between the functions, we write:\[f(x) - g(x) = \frac{2}{x+1} - \frac{x}{x+1}\]With the same denominator, subtract the numerators:\[f(x) - g(x) = \frac{2 - x}{x+1}\]The domain is all real numbers except \(x = -1\).

Step 3: Find fg

To multiply the functions, we multiply the numerators and the denominators:\[f(x) \cdot g(x) = \frac{2}{x+1} \cdot \frac{x}{x+1} = \frac{2x}{(x+1)^2}\]The domain is all real numbers except \(x = -1\).

Step 4: Find f/g

To divide the functions, we write:\[\frac{f(x)}{g(x)} = \frac{\frac{2}{x+1}}{\frac{x}{x+1}}\]This is equivalent to:\[\frac{f(x)}{g(x)} = \frac{2}{x+1} \cdot \frac{x+1}{x} = \frac{2}{x}\]The domain excludes \(x = 0\) and \(x = -1\).

Step 5: Summarize Domains

The domains for each operation are:- \(f+g\), \(f-g\), and \(fg\) are undefined at \(x = -1\).- \(f/g\) is undefined at \(x = -1\) and \(x = 0\).Write answers as intervals:- \(f+g, f-g, fg\) : \((-\infty, -1) \cup (-1, \infty)\)- \(f/g\) : \((-\infty, -1) \cup (-1, 0) \cup (0, \infty)\)

Key Concepts

Domain

The domain of a function is the set of all possible input values (x-values) that the function can accept without breaking any mathematical rules. For rational functions like \(f(x) = \frac{2}{x+1}\) and \(g(x) = \frac{x}{x+1}\), we need to ensure the denominator is not zero because division by zero is undefined.
For the example functions given, both share the same denominator \((x+1)\). Therefore, both functions are undefined at \(x = -1\). When performing operations like addition, subtraction, multiplication, or division, we have to consider the domains of both functions being combined. This means:

• Addition, subtraction, or multiplication: exclude \(x=-1\).
• Division: exclude where either denominator is zero; here, \(x=-1\) and \(x=0\).

To summarize:

• For \(f+g\), \(f-g\), and \(fg\): domain is \((-\infty, -1) \cup (-1, \infty)\).
• For \(f/g\): domain is \((-\infty, -1) \cup (-1, 0) \cup (0, \infty)\).

Addition of Functions

Adding functions involves combining them by summing their output values for any input from their common domain. For functions \(f(x) = \frac{2}{x+1}\) and \(g(x) = \frac{x}{x+1}\), their addition \(f(x) + g(x)\) simplifies nicely due to having the same denominator:
\[f(x) + g(x) = \frac{2}{x+1} + \frac{x}{x+1} = \frac{2+x}{x+1}\]Because \(x = -1\) would make the denominator zero, it's not part of the domain for this addition. Thus, the domain of \(f+g\) is all real numbers except \(x=-1\). This means the combined function can accept any real x-value except where division by zero occurs, specifically at \(x=-1\). This ensures no undefined operations when evaluating the function.

Subtraction of Functions

Subtraction of functions is similar to addition, but instead you subtract their outputs. Using the same functions \(f(x)\) and \(g(x)\), for their subtraction \(f(x) - g(x)\), since the denominators match, you directly subtract the numerators:
\[f(x) - g(x) = \frac{2}{x+1} - \frac{x}{x+1} = \frac{2-x}{x+1}\]The requirement remains to avoid making the denominator zero, so \(x = -1\) is still excluded from the domain. The domain of \(f-g\) is therefore the same as in addition: all real numbers except \(x = -1\). It helps us combine functions safely by specifying allowable input values.

Multiplication of Functions

In multiplication, you multiply the output values of the two functions. With \(f(x)\) and \(g(x)\), multiplying them results in:
\[f(x) \cdot g(x) = \frac{2}{x+1} \cdot \frac{x}{x+1} = \frac{2x}{(x+1)^2}\]The domain remains excluding \(x = -1\), following the same rule about not allowing the denominator to become zero. While the expression and its structure change during multiplication, it's important to remember: the domain constraints carry over. Thus, the domain of \(fg\) must exclude any x-values that appear in the denominator of either function.

Division of Functions

Division differs from other operations because you divide one function's output by another. This means not only should the denominator in the division itself not be zero, but also that of the original functions. Given our functions:
\[\frac{f(x)}{g(x)} = \frac{\frac{2}{x+1}}{\frac{x}{x+1}} = \frac{2}{x}\]Here, both zeroes of the original denominators and places where \(x\) makes the result undefined (like \(x=0\)) are excluded:

• \(x = -1\) makes \(x+1=0\), causing undefined conditions in both \(f\) and \(g\).
• \(x = 0\) makes the simplified \(\frac{2}{x}\) undefined.

Thus, the domain of \(f/g\) excludes both \(x = -1\) and \(x = 0\), highlighting the need to meticulously track all potential points of undefined behavior.