Problem 66 - College Algebra Essentials

Question

Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=\frac{2}{x}, g(x)=\frac{2}{x}$$

Step-by-Step Solution

Answer

\((f \circ g)(x) = x\), \((g \circ f)(x) = x\), \((f \circ g)(2) = 2\) and \((g \circ f)(2) = 2\)

Step 1: Compose functions

To find \((f \circ g)(x)\), apply function \(g(x)\) first, then \(f(x)\). So: \[(f \circ g)(x) = f[g(x)] = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x\] Similar steps are followed to solve \((g \circ f)(x)\), as the functions are the same, the output remains \(x\).

Step 2: Evaluate Composed Functions

After having computed the composed functions, we can substitute \(x = 2\) into the equation, hence: \[(f \circ g)(2) = 2 \quad \text{and} \quad (g \circ f)(2) = 2 \] since the output of the composed functions is \(x\).

Key Concepts

Inverse Functions

The idea behind inverse functions is that they reverse the effect of a function. Often, students initially struggle to grasp this concept, but breaking it down can help. In simple terms, if you have a function \( y = f(x) \), its inverse \( f^{-1}(x) \) satisfies the equation \( x = f^{-1}(y) \). This means applying a function followed by its inverse will get you back to where you started.
In the exercise, both \( f(x) = \frac{2}{x} \) and \( g(x) = \frac{2}{x} \) primarily act as their own inverse — if you apply either function after the other, they effectively cancel each other out, giving you back the original input \( x \). This showcases how inverses work in function composition, where done sequentially, they lead back to the initial value.

Function Evaluation

Function evaluation involves plugging a specific value into a function to get a result. It’s like following the recipe in a cookbook - you input a particular \( x \) and get the output after processing it with the function formula. In the exercise, students encounter function evaluation when they need to determine the outcome of composed functions at \( x = 2 \).
The calculation \((f \, \circ \, g)(2)\) involves substituting \( x = 2 \) into the function \( g(x) \) first, then using that result in \( f(x) \). As shown in the solution, since \((f \, \circ \, g)(x)\) returns \( x \), substituting \( 2 \) back in keeps that result unchanged. Practicing function evaluation helps solidify one’s understanding of how inputs are transformed into outputs through a mathematical function.

Algebraic Functions

Algebraic functions are constructed using algebraic operations such as addition, subtraction, multiplication, and division. They form the building blocks of higher mathematical concepts. The exercise deals with simple algebraic functions, \( f(x) = \frac{2}{x} \) and \( g(x) = \frac{2}{x} \), showcasing operations often encountered in algebra.

Key characteristics include:

• Processing inputs via regular mathematical operations.
• Utilizing expressions like fractions or polynomials.
• Frequent use in more complex equations and their transformations.

Recognizing algebraic structures like these helps students understand more complex problems as they progress. Simplifying and composing these functions, as shown in the exercise, builds foundational skills that are essential for advanced study.