Question

Show that \(f\) and \(g\) are inverses of each other by verifying that \(f[g(x)]=x\) and \(g[f(x)]=x\). $$ f(x)=\frac{1+x}{1-x} ; \quad g(x)=\frac{x-1}{x+1} $$

Short answer

We have the functions \(f(x) = \frac{1+x}{1-x}\) and \(g(x) = \frac{x-1}{x+1}\). To show that they are inverses, we computed and simplified \(f[g(x)]\) and \(g[f(x)]\). After simplification, both expressions resulted in \(x\), proving that \(f(x)\) and \(g(x)\) are inverses of each other.

Step-by-step solution

Find f[g(x)]

To find f[g(x)], we will substitute g(x) into the function f(x) where x is. Since \(g(x) = \frac{x-1}{x+1}\), we substitute this into the f(x) expression: \(f[g(x)] = f\left(\frac{x-1}{x+1}\right) = \frac{1+\frac{x-1}{x+1}}{1-\frac{x-1}{x+1}}\)

Simplify f[g(x)]

Now, we will simplify the expression for f[g(x)]: \( f[g(x)] = \frac{\frac{x+1+x-1}{x+1}}{\frac{x+1-x+1}{x+1}} \\ = \frac{\frac{2x}{x+1}}{\frac{2}{x+1}} \) To simplify further, we can cancel out the common factors in the numerator and denominator: \( f[g(x)] = \frac{2x}{x+1} \cdot \frac{x+1}{2} \\ = x \) Thus, we have shown that f[g(x)] = x.

Find g[f(x)]

Now, we will find g[f(x)] by substituting f(x) into the function g(x) where x is. Since \(f(x) = \frac{1+x}{1-x}\), we substitute this into the g(x) expression: \(g[f(x)] = g\left(\frac{1+x}{1-x}\right) = \frac{\left(\frac{1+x}{1-x}\right) -1}{\left(\frac{1+x}{1-x}\right) +1}\)

Simplify g[f(x)]

Now, we will simplify the expression for g[f(x)]: \( g[f(x)] = \frac{\frac{1+x-1+x}{1-x}}{\frac{1+x+1-x}{1-x}}\\ = \frac{\frac{2x}{1-x}}{\frac{2}{1-x}} \) To simplify further, we can cancel out the common factors in the numerator and denominator: \( g[f(x)] = \frac{2x}{1-x} \cdot \frac{1-x}{2} \\ = x \) Thus, we have shown that g[f(x)] = x. Since we have verified that f[g(x)] = x and g[f(x)] = x, we can conclude that the functions f(x) and g(x) are inverses of each other.

Key concepts

Calculus

Calculus, at its core, involves studying change and motion. One aspect that frequently arises is the interplay between functions and their inverses. Inverse functions reverse the effect of the original function. In the case of calculus, we often deal with derivatives and integrals of inverse functions. However, to tackle derivatives or integrals, we need to first understand the functions themselves.

For example, if we have a function that takes an input and squares it, the inverse function would take the square and return the original input. When it comes to verifying inverse functions, calculus focuses on ensuring that this 'undoing' process is valid. The process involves composing functions with their proposed inverses and simplifying to check if the result is the identity function, signifying an undone action.

Function Composition

Understanding Function Composition

Function composition is the process of applying one function to the results of another. The notation \(f \circ g\)(x), or \(f[g(x)]\), represents the output of function \(f\) after \(g\) has been applied to \(x\).

In the inverse verification process, this concept is essential as it demonstrates whether two functions are truly inverses. By composing \(f(x)\) with \(g(x)\) and vice versa, and subsequently simplifying, we aim for the result to be \(x\) itself. If successful, the composition effectively undoes the action of its counterpart, which is how we recognize inverse functions.

Applying this to the exercise provided, we observed \(f[g(x)]\) and \(g[f(x)]\) through their respective compositions. Each composition simplified to \(x\), affirming that \(f\) and \(g\) undo each other's actions - a definitive characteristic of inverse functions.

Simplifying Expressions

The Power of Simplification

Simplifying expressions is a fundamental tool in mathematics that aids in understanding and solving problems more efficiently. The exercise involves transforming the composition of functions into a simpler form to reveal an underlying identity or relationship - in this case, that \(f\) and \(g\) are inverses.

In practice, simplifying can involve combining like terms, cancelling out common factors, or applying algebraic rules to reduce complexity. As shown in the exercise, after composing the functions, we simplified the complex fractions step by step, eliminating common factors that appear in both the numerator and the denominator. The skill of simplifying not only provides clarity but also supports the verification process by illustrating that the composition of the functions indeed results in the identity function \(x\).

This concept is fundamental in calculus and algebra alike and is particularly reflected in exercises that involve function compositions and inverse function verification. Simplification allows us to conclude confidently that \(f\) and \(g\) are, without a doubt, inverse functions.