Problem 35 - Algebra and Trigonometry

Question

Use the Inverse Function Property to show that \(f\) and \(g\) are inverses of each other. $$ f(x)=\frac{x+2}{x-2} ; \quad g(x)=\frac{2 x+2}{x-1} $$

Step-by-Step Solution

Answer

The functions \(f(x)\) and \(g(x)\) are inverses of each other.

Step 1: Understanding the inverse function property

Two functions, \(f\) and \(g\), are inverses if \(f(g(x)) = x\) and \(g(f(x)) = x\) for all \(x\) in the domains of \(g\) and \(f\) respectively. We'll use this property to show that \(f(x)\) and \(g(x)\) are inverses of each other.

Step 2: Calculate \(f(g(x))\)

First, calculate \(f(g(x))\):\[f(g(x)) = f\left(\frac{2x+2}{x-1}\right) = \frac{\frac{2x+2}{x-1} + 2}{\frac{2x+2}{x-1} - 2}\]Simplify the expression:\[= \frac{\frac{2x+2}{x-1} + \frac{2(x-1)}{x-1}}{\frac{2x+2}{x-1} - \frac{2(x-1)}{x-1}} = \frac{\frac{2x+2 + 2x - 2}{x-1}}{\frac{2x+2 - 2x + 2}{x-1}} \]\[= \frac{\frac{4x}{x-1}}{\frac{4}{x-1}} = x\]

Step 3: Calculate \(g(f(x))\)

Next, calculate \(g(f(x))\):\[g(f(x)) = g\left(\frac{x+2}{x-2}\right) = \frac{2\left(\frac{x+2}{x-2}\right) + 2}{\frac{x+2}{x-2} - 1}\]Simplify the expression:\[= \frac{\frac{2(x+2)}{x-2} + \frac{2(x-2)}{x-2}}{\frac{x+2}{x-2} - \frac{x-2}{x-2}} = \frac{\frac{2x+4 + 2x - 4}{x-2}}{\frac{x+2 - x + 2}{x-2}} \]\[= \frac{\frac{4x}{x-2}}{\frac{4}{x-2}} = x\]

Step 4: Conclusion

Since \(f(g(x)) = x\) and \(g(f(x)) = x\), the functions \(f(x)\) and \(g(x)\) are inverses of each other based on the inverse function property.

Key Concepts

Inverse Function Property

The inverse function property is a unique characteristic that helps identify when two functions essentially "undo" each other. For two functions, \(f\) and \(g\), to be inverses, they must satisfy the following conditions:

• The composition \(f(g(x))\) must simplify to \(x\).
• The composition \(g(f(x))\) must simplify to \(x\).

This means that applying function \(g\) and then function \(f\) gives you your original input \(x\), and vice versa. This property is useful in solving problems where we need to determine if one function is an inverse of another. When these compositions simplify to \(x\), it proves the functions are inversely related.

Function Composition

Function composition is like chaining functions together. Imagine feeding the output of one function as the input to another. For example, if you have two functions \(f(x)\) and \(g(x)\), the composition \(f(g(x))\) means you first apply \(g\) to \(x\), then apply \(f\) to the result.

A simple illustration is when you put a number through \(g\), and the result goes through \(f\). This chain reaction determines whether these two functions are inverses. In the original exercise, the compositions \(f(g(x))\) and \(g(f(x))\) were calculated to establish their inverses, helping to determine that these functions bring us back to our starting value \(x\).

Algebraic Simplification

Algebraic simplification is the process of breaking down complex expressions into simpler forms. It is a crucial step in proving that functions are inverses, as it often involves rationalizing numerators and denominators or combining like terms to reduce expressions.

In the step-by-step solution, simplifying the compositions \(f(g(x))\) and \(g(f(x))\) involved breaking fractions into components that could be combined or canceled out. The goal was to reach the expression \(x\), showcasing the expected outcome of the inverse function property through concise algebraic steps. Mastering these simplifications is a key algebraic skill for solving such problems.

Domain of a Function

The domain of a function refers to all possible input values (often \(x\)-values) for which the function is defined. When dealing with inverse functions, understanding each function's domain is essential to avoid undefined expressions, such as division by zero.

In this exercise, we must consider the domains of \(f(x)\) and \(g(x)\) to ensure the functions and their inverses do not result in undefined forms. For example, for \(f(x) = \frac{x+2}{x-2}\), \(x\) cannot be 2, as it would cause division by zero. Similarly, precise checking of function domains ensures that the inverse compositions \(f(g(x))\) and \(g(f(x))\) remain valid and contribute accurately to verifying the inverse function property.