Problem 66 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

a. \((f \circ g)(x) = x\), b. \((g \circ f)(x) = x\), c. \((f \circ g)(2) = 2\), d. \((g \circ f)(2) = 2\)

Step 1: Find \((f \circ g)(x)\)

The composition function \(f \circ g\) means to apply the function \(g(x)\) first and then apply the function \(f(x)\) to the result. Write it as \(f(g(x))\). In this case, since \(f(x) = \frac{2}{x}\) and \(g(x) = \frac{2}{x}\), then \(f(g(x)) = f(\frac{2}{x}) = \frac{2}{2/x} = x.

Step 2: Find \((g \circ f)(x)\)

Similarly, the composition function \(g \circ f\) means to apply the function \(f(x)\) first and then apply the function \(g(x)\) to the result. Write it as \(g(f(x))\). In this case, since \(g(x) = \frac{2}{x}\) and \(f(x) = \frac{2}{x}\), then \(g(f(x)) = g(\frac{2}{x}) = \frac{2}{2/x} = x.

Step 3: Find \((f \circ g)(2)\)

To find \((f \circ g)(2)\), substitute \(x = 2\) into the result of \((f \circ g)(x)\). It gives \(f(g(2)) = 2.\)

Step 4: Find \((g \circ f)(2)\)

Similarly, to find \((g \circ f)(2)\), substitute \(x = 2\) into the result of \((g \circ f)(x)\). It gives \(g(f(2)) = 2.

Key Concepts

Inverse Functions

Inverse functions are pairs of functions that "undo" each other. When you apply an inverse function to an original function, they cancel each other out, leaving you with the initial value. For example, if you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), then applying \( f^{-1} \) to \( f(x) \) will return \( x \). Such that it can be expressed as:\[ f(f^{-1}(x)) = x \ \text{and} \ f^{-1}(f(x)) = x \]In the exercise, both functions \( f(x) \) and \( g(x) \) are \( \frac{2}{x} \), showing how each function can act as an inverse. When \((f \circ g)(x)\) yields \(x\) (the original input), \(f(x)\) and \(g(x)\) effectively act as inverses in this context. Highlighting this "un-doing" feature showcases their role as reverse operations of each other. It's an important concept, because understanding inverses can deeply enhance algebraic problem-solving skills.
Thus, one significant trait of this exercise is comprehending how function reversals work, even when they involve fractions or other operations as in our example.

Substitution

Substitution in mathematics is a technique where one replaces a variable with a given value or another expression. This is particularly useful for simplification and evaluation. In the context of functions, substitution allows for the evaluation of composite functions like \((f \circ g)(x)\) or \((g \circ f)(x)\).
In this problem, substitution is used to first solve the inside function before handling the outer one. For example, obtaining \((f \circ g)(x)\) means substituting \(g(x) = \frac{2}{x}\) into \(f\), leading to:\[ f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{2/x} = x \]When evaluating these compositions with a specific number, like \(2\), substitution helps find precise answers, such as in \((f \circ g)(2)\). It is all about substituting \(x\) with \(2\), giving:\[ f(g(2)) = 2 \]This problem emphasizes that substitution is crucial for solving and simplifying compositions in function operations, allowing for systematic approaches to finding values and understanding behavior in algebraic functions.

Identity Function

An identity function is a type of function in mathematics which returns exactly the same value that is input. Denoted as \( I(x) = x \), it serves as a neutral element in function composition. When composing any function with an identity function, the original function is unaffected, such that:\[ f(I(x)) = f(x) \ \text{and} \ I(f(x)) = f(x) \]In this exercise, both \((f \circ g)(x)\) and \((g \circ f)(x)\) resulted in \(x\). This relationship reflects the property of the identity function as these compositions mimic it.
The significance of recognizing identity functions lies in verifying a lack of change through composition. In algebra, identifying such situations helps ensure mathematical operations were executed correctly and provides deep insight into function behaviors. Understanding identity functions assists in predicting the outcomes of function compositions and verifying solutions, affirming the correctness of problem-solving strategies. This adds another layer of confidence when dealing with complex function expressions as students progress in mathematics.