91Ó°ÊÓ

Function composition involves applying one function to the result of another function. In this exercise, we want to check if two functions are inverses. If they are, then composing them in any order should return the input variable, say, \(x\). This means we need to find \(f(g(x))\) and \(g(f(x))\) and see if both are equal to \(x\).

For example, if you have \(f(x)\) and \(g(x)\) as functions, then:

• First, substitute \(g(x)\) into \(f\):
  \[ f(g(x)) = f \left( \frac{1-x}{x} \right) \]
• Next, simplify the expression to see if it equals \(x\).
• Do the same for \(g(f(x))\).

This process helps in determining whether every operation by \(f\) can be undone by \(g\), and vice versa.

Algebraic manipulation involves using algebraic methods to simplify or rearrange expressions. When checking whether two functions are inverses, it’s essential to skillfully manipulate the equations:

For instance, if you have the composed function \(f(g(x))\):

• Start with \( f(g(x)) = f \left( \frac{1-x}{x} \right) \)
• Follow through with substitution: \[ f \left( \frac{1-x}{x} \right) = \frac{-1}{\frac{1-x}{x} + 1} \]
• Simplify this by combining fractions: \[ = \frac{-1}{\frac{1-x + x}{x}} = \frac{-1}{\frac{1}{x}} = -x \]

Unfortunately, \(-x eq x\); hence, we conclude that \(f\) and \(g\) are not inverses.
Practicing such steps regularly improves your skills in handling complex function compositions.

Function properties play a key role in understanding inverse functions. Here are a few critical properties to consider:

1. One-to-One Property: A function must be one-to-one (bijective) to have an inverse. If a function maps different inputs to the same output, its inverse won't be well-defined.
2. Identity Property: For \(f\) and \(g\) to be inverses, it must hold that \(f(g(x)) = x\) and \(g(f(x)) = x\). This means each function exactly reverses the effect of the other.
3. Domain and Range: The domain of \(f\) must match the range of \(g\) and vice versa.

Let's evaluate \( g(f(x)) \) for confirmation:

• Substitute \( f(x) \) into \( g \):
  \[ g(f(x)) = g \left( \frac{-1}{x+1} \right) \]
• Simplify further: \[ = \frac{1 - \left( \frac{-1}{x+1} \right)}{\frac{-1}{x+1}} = \frac{ 1 + \frac{1}{x+1} }{ \frac{-1}{x+1} } = \left(1 + \frac{1}{x+1}\right) \times (-(x+1)) = -(x+2) \]

Again, \(-(x+2) eq x\), confirming that \( f \) and \( g \) are not inverses of each other. Understanding these properties helps you grasp why certain functions can't be inverses and strengthens your foundation in algebra.