Problem 56 - Intermediate Algebra

Question

Show that each pair of functions are inverses. $$ f(x)=5 x-1, f^{-1}(x)=\frac{x+1}{5} $$

Step-by-Step Solution

Answer

The functions are inverses because \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).

Step 1: Apply the Definition of Inverse Functions

Two functions, \(f\) and \(f^{-1}\), are inverses if \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) for all \(x\) in the domains of \(f^{-1}\) and \(f\) respectively. We need to verify both these conditions.

Step 2: Verify \(f(f^{-1}(x)) = x\)

Start by substituting \(f^{-1}(x)\) into \(f\). Given \(f(x) = 5x - 1\) and \(f^{-1}(x) = \frac{x+1}{5}\), calculate \(f(f^{-1}(x))\):\[f\left(\frac{x+1}{5}\right) = 5\left(\frac{x+1}{5}\right) - 1\]Simplify:\[= x + 1 - 1 = x\]Thus, \(f(f^{-1}(x)) = x\).

Step 3: Verify \(f^{-1}(f(x)) = x\)

Now substitute \(f(x)\) into \(f^{-1}\). Using the equations given, compute \(f^{-1}(f(x))\):\[f^{-1}(5x - 1) = \frac{(5x - 1) + 1}{5}\]Simplify:\[= \frac{5x}{5} = x\]Thus, \(f^{-1}(f(x)) = x\).

Step 4: Conclusion

Both conditions \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) are satisfied. This confirms that \(f(x) = 5x - 1\) and \(f^{-1}(x) = \frac{x+1}{5}\) are indeed inverse functions.

Key Concepts

Function Operations

When dealing with functions, one of the crucial concepts is understanding how to perform operations with them. In mathematics, functions can be combined and manipulated through various operations. The main operations involve addition, subtraction, multiplication, and division of functions. However, in the context of inverse functions, the primary operation of interest is function composition.

Function composition involves merging two functions such that the output of one function becomes the input of another. It is represented as \((f \circ g)(x) = f(g(x))\).

• In our exercise, to prove that two functions are inverses, we use function composition.
• The compositions needed are \(f(f^{-1}(x))\) and \(f^{-1}(f(x))\).

These compositions check if applying one function after the other results in the identity function, essentially returning the original input, \(x\). This process highlights the reversed operation that inverse functions perform on each other.

Verification of Inverses

Verifying if two functions are inverses involves a straightforward process. The definition of inverse functions states that for a pair of functions \(f\) and \(f^{-1}\), both conditions \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) must hold true across their domains.

To verify, start by substituting the inverse function into the original function.

• Given \(f(x) = 5x - 1\) and \(f^{-1}(x) = \frac{x+1}{5}\), replace \(f^{-1}(x)\) in \(f(x)\), resulting in
  \[f\left(\frac{x+1}{5}\right) = 5\left(\frac{x+1}{5}\right) - 1\], which simplifies to \(x\).

Next, substitute the original function into the inverse function:

• Using \(f^{-1}(f(x))\), plug \(f(x)\) into \(f^{-1}\), leading to
  \[f^{-1}(5x - 1) = \frac{(5x - 1) + 1}{5}\], which also simplifies to \(x\).

Both results leading to \(x\) confirm that \(f\) and \(f^{-1}\) indeed act as inverse functions, as expected.

Algebraic Functions

Algebraic functions encompass a wide variety of functions formed by combining algebraic operations like addition, subtraction, multiplication, division, and taking roots. Linear functions, polynomials, and rational functions all fall under this umbrella.

In the problem presented, \(f(x) = 5x - 1\) represents a linear function, while \(f^{-1}(x) = \frac{x+1}{5}\) is a rational function. Linear functions are among the simplest algebraic functions and involve expressions of the form \(ax + b\), where \(a\) and \(b\) are constants.

The clever manipulation of these functions through algebraic methods—such as solving equations and simplifying expressions—allows us to find inverses and perform operations efficiently.

• For instance, if you solve the equation \(y = 5x - 1\) for \(x\), you'll derive the inverse function \(x = \frac{y + 1}{5}\), which is rearranged as \(f^{-1}(x)\).
• This demonstrates how algebraic techniques empower us to not only manage but also solve complex functional equations.