Problem 44 - Algebra and Trigonometry

Question

Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$ f(x)=\frac{2}{x}, \quad g(x)=\frac{x}{x+2} $$

Step-by-Step Solution

Answer

The compositions are: \(f \circ g = \frac{2(x+2)}{x}\), \(g \circ f = \frac{1}{x+1}\), \(f \circ f = x\), \(g \circ g = \frac{x}{x+6}\). Their domains exclude values making denominators zero.

Step 1: Determine f(g(x))

To find the composition function \(f \circ g(x)\), we substitute \(g(x)\) into \(f(x)\). Given \(g(x) = \frac{x}{x+2}\), substitute into \(f\):\[ f(g(x)) = f\left( \frac{x}{x+2} \right) = \frac{2}{\frac{x}{x+2}} = \frac{2(x+2)}{x} \]Thus, the function \(f \circ g(x) = \frac{2(x+2)}{x}\). The domain of \(f \circ g\) is all real numbers except \(x = -2\) and \(x = 0\) since these make the denominator zero.

Step 2: Determine g(f(x))

For \(g \circ f(x)\), we need to substitute \(f(x)\) into \(g(x)\). Since \(f(x) = \frac{2}{x}\), substitute into \(g\):\[ g(f(x)) = g\left( \frac{2}{x} \right) = \frac{\frac{2}{x}}{\frac{2}{x} + 2} = \frac{\frac{2}{x}}{\frac{2+2x}{x}} = \frac{2}{2+2x} \]Thus, \(g \circ f(x) = \frac{1}{x+1}\). The domain of \(g \circ f\) is all real numbers except \(x = 0\) and \(x = -1\).

Step 3: Determine f(f(x))

To find \(f \circ f(x)\), substitute \(f(x)\) into itself:\[ f(f(x)) = f\left( \frac{2}{x} \right) = \frac{2}{\frac{2}{x}} = x \]Therefore, \(f \circ f(x) = x\). The domain of \(f \circ f\) is all real numbers except \(x = 0\), as this would make \(f(x)\) undefined.

Step 4: Determine g(g(x))

For \(g \circ g(x)\), substitute \(g(x)\) into itself:\[ g(g(x)) = g\left( \frac{x}{x+2} \right) = \frac{\frac{x}{x+2}}{\frac{x}{x+2} + 2} = \frac{\frac{x}{x+2}}{\frac{x+2}{x+2} + \frac{4}{x+2}} = \frac{x}{x + 6} \]Thus, \(g \circ g(x) = \frac{x}{x+6}\). The domain is all real numbers except \(x = -2\) and \(x = -6\).

Key Concepts

Domain of Functions

The domain of a function is an essential concept in mathematics. It refers to all the possible values of the input (or the variable) that allows the function to work without any issues.
In essence, it's about determining which numbers one can substitute in place of 'x' that will yield a valid output.

• If you have a function like a fraction, you need to ensure the denominator never becomes zero. Otherwise, it would be undefined.
• For example, in the function \( f(x)=\frac{2}{x} \), any value except zero is acceptable for 'x' because it would make the fraction undefined if \( x=0 \).
• Similarly, for a rational function like \( g(x)=\frac{x}{x+2} \), the domain excludes any 'x' that makes \( x+2 \) zero, which is \( x = -2 \).

This reflects the importance of always checking the denominator when dealing with rational functions to ensure valid domains.

Rational Functions

A rational function is a type of function that's expressed as the ratio of two polynomials. In simpler terms, it takes the form \( \frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) eq 0 \).

• These functions are quite common and have several real-world applications ranging from physics to economics.
• If you look at our examples like \( f(x) = \frac{2}{x} \) and \( g(x) = \frac{x}{x+2} \), you can see they both fit the pattern of rational functions.
• The main rule to remember with rational functions is the denominator must not be zero, or the function becomes undefined.

Rational functions can exhibit interesting properties, such as vertical asymptotes, at the values of 'x' where the denominator zeroes out. Understanding these functions helps in analyzing their behavior on graphs.

Algebraic Operations

Algebraic operations involve using arithmetic operations (addition, subtraction, multiplication, and division) with functions. When working with function compositions such as \( f \circ g \) or \( g \circ f \), you're essentially plugging one function into another, which often involves algebraic manipulation.
This involves replacing the variable in one function with another function entirely and simplifying the result.

• For example, to find \( f \circ g(x) \), substitute \( g(x) = \frac{x}{x+2} \) into \( f(x) \) resulting in \( \frac{2}{\frac{x}{x+2}} \), which simplifies to \( \frac{2(x+2)}{x} \).
• To further simplify or manipulate these expressions, you may need to perform operations like multiplying by the reciprocal or factoring.
• This process is essential for understanding how functions interact when composed and how their domains are determined from such operations.

Can't forget to simplify the results! Algebraic rules will always guide how we evaluate and reduce our expressions.