Problem 33 - Algebra and Trigonometry

Question

Use the Inverse Function Property to show that \(f\) and \(g\) are inverses of each other. $$ f(x)=\frac{1}{x-1}, \quad x \neq 1 ; \quad g(x)=\frac{1}{x}+1, \quad x \neq 0 $$

Step-by-Step Solution

Answer

Yes, \(f(x)\) and \(g(x)\) are inverses.

Step 1: Understand the Inverse Function Property

The Inverse Function Property states that if function \(f\) and function \(g\) are inverses, then \(f(g(x)) = x\) for all \(x\) in the domain of \(g\), and \(g(f(x)) = x\) for all \(x\) in the domain of \(f\). This means that applying \(g\) and then \(f\) (or vice versa) should return the original value \(x\).

Step 2: Check \(f(g(x)) = x\)

Compute \(f(g(x))\) where \(g(x) = \frac{1}{x} + 1\). Substituting into \(f(x)\): \[ f(g(x)) = f\left(\frac{1}{x} + 1\right) = \frac{1}{\left(\frac{1}{x} + 1\right) - 1} = \frac{1}{\frac{1}{x}} = x. \] Thus, for \(x eq 0\), \(f(g(x)) = x\).

Step 3: Check \(g(f(x)) = x\)

Compute \(g(f(x))\) where \(f(x) = \frac{1}{x-1}\). Substituting into \(g(x)\): \[ g(f(x)) = g\left(\frac{1}{x-1}\right) = \frac{1}{\left(\frac{1}{x-1}\right)} + 1 = (x-1) + 1 = x. \] Thus, for \(x eq 1\), \(g(f(x)) = x\).

Step 4: Conclusion

Since both \(f(g(x)) = x\) for \(x eq 0\) and \(g(f(x)) = x\) for \(x eq 1\), function \(f\) and function \(g\) are inverses of each other by the Inverse Function Property.

Key Concepts

Function Composition

Function composition is essentially the process of combining two functions to form a new function. Think of it as one function being the input to another function.
For example, in the given exercise, we are looking at functions \(f(x)\) and \(g(x)\). The composition of these functions is represented by \(f(g(x))\) and \(g(f(x))\).
The idea is to substitute the output of one function (\(g(x)\), for example) into another function (\(f(x)\) in this case) to see what happens.
Function composition is a handy tool in mathematics because it helps us understand how different functions interact with each other. It's significant in verifying inverse functions because it should bring us back to the original input \(x\). That's the core of the Inverse Function Property: if applying a sequence of functions (and their inverses correctly), we end up with our starting point.

Domain of a Function

The domain of a function is the set of all possible inputs for which the function is defined. It is crucial to determine the domain to ensure that a function doesn't have undefined values.
In our example, for function \(f(x) = \frac{1}{x-1}\), the domain excludes \(x = 1\) because this would make the denominator zero, leading to an undefined expression.
Similarly, for \(g(x) = \frac{1}{x} + 1\), the domain excludes \(x = 0\) for the same reason — division by zero is undefined.
Understanding the domain of the functions is important when checking if they are inverses of each other. Functions can only be inverses if they are invertible for their entire domain.

Inverse Functions

Inverse functions essentially undo the action of each other. Mathematically, for two functions \(f\) and \(g\) to be inverses, they must satisfy two conditions:

• \(f(g(x)) = x\) for every \(x\) in the domain of \(g\)
• \(g(f(x)) = x\) for every \(x\) in the domain of \(f\)

In the example provided, computations for both \(f(g(x))\) and \(g(f(x))\) give us \(x\), verifying that each function undoes the effect of the other.
It’s like having a lock and a key. \(f\) is the lock, \(g\) is the key, and using them together as intended gets you back to the unlocked position—'back to \(x\)', as represented in math.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions in mathematics to reveal underlying properties.
In verifying inverse functions like \(f\) and \(g\), algebraic manipulation is used to simplify function compositions, such as \(f(g(x))\) and \(g(f(x))\), to confirm if they equal \(x\).
For instance, substituting \(g(x) = \frac{1}{x} + 1\) into \(f(x) = \frac{1}{x-1}\) requires careful manipulation of fractions and like terms to result in \(x\).
These steps involve simplifying complex fractions and adding terms correctly, which is fundamental for making sure you accurately determine if two functions are inverses.