Problem 21 - Algebra for College Students

Question

Verify that the two given functions are inverses of each other. $$ f(x)=-\frac{1}{2} x+\frac{5}{6} \text { and } g(x)=-2 x+\frac{5}{3} $$

Step-by-Step Solution

Answer

Yes, they are inverses.

Step 1: Find Composition of Functions

To verify that two functions are inverses, first find the composition of the functions. We will first compute \( f(g(x)) \). Substitute \( g(x) = -2x + \frac{5}{3} \) into \( f(x) \).

Step 2: Calculate f(g(x))

Substitute \( g(x) = -2x + \frac{5}{3} \) into \( f(x) = -\frac{1}{2}x + \frac{5}{6} \):\[ f(g(x)) = f\left(-2x + \frac{5}{3}\right) \]\[ = -\frac{1}{2}(-2x + \frac{5}{3}) + \frac{5}{6} \]\[ = x - \frac{5}{6} + \frac{5}{6} \]\[ = x \]Indicating that \( f(g(x)) = x \).

Step 3: Find Composition of Functions in Reverse

Next, compute \( g(f(x)) \). Substitute \( f(x) = -\frac{1}{2}x + \frac{5}{6} \) into \( g(x) \).

Step 4: Calculate g(f(x))

Substitute \( f(x) = -\frac{1}{2}x + \frac{5}{6} \) into \( g(x) = -2x + \frac{5}{3} \):\[ g(f(x)) = g\left(-\frac{1}{2}x + \frac{5}{6}\right) \]\[ = -2\left(-\frac{1}{2}x + \frac{5}{6}\right) + \frac{5}{3} \]\[ = x - \frac{5}{3} + \frac{5}{3} \]\[ = x \]This shows that \( g(f(x)) = x \).

Step 5: Conclude Inverse Relationship

Since both compositions \( f(g(x)) \) and \( g(f(x)) \) result in \( x \), the functions \( f(x) \) and \( g(x) \) are inverses of each other.

Key Concepts

Composition of Functions

Composition of functions is a fascinating concept in mathematics. It involves applying one function to the result of another, ultimately forming a single new function. This concept is often represented by two functions, say \( f(x) \) and \( g(x) \). The composition \( f(g(x)) \) means plug the output of \( g \) into \( f \).
Let's illustrate this with our functions from the exercise, where \( f(x) = -\frac{1}{2}x + \frac{5}{6} \) and \( g(x) = -2x + \frac{5}{3} \). When we perform the composition \( f(g(x)) \), we substitute \( g(x) \) into \( f \). This process involves replacing every instance of \( x \) in \( f(x) \) with \( -2x + \frac{5}{3} \) from \( g(x) \).

• First substitute: \( f(g(x)) = f\left(-2x + \frac{5}{3}\right) \)
• Then compute: \( = -\frac{1}{2}(-2x + \frac{5}{3}) + \frac{5}{6} \)
• Result: \( = x \)

Getting \( x \) confirms that these function operations bring us back to the input, suggesting a special relationship between \( f \) and \( g \).
The interesting thing about function composition and inverse operations is its symmetry. Like a puzzle piece fitting perfectly within another.

Mathematical Verification

A core part of working with functions is verifying their properties, such as determining if they are inverses. To say that two functions are inverses means they "undo" each other's operations. The composition of two inverses, \( f(g(x)) \) or \( g(f(x)) \), should result in the identity function, \( x \).
In our exercise, after calculating both \( f(g(x)) = x \) and \( g(f(x)) = x \), we've mathematically verified that these functions are indeed inverses. Here's how it unfolds:

• First, when you compute \( f(g(x)) \), you get your original input back.
• Then, repeating this with \( g(f(x)) \) also returns \( x \).

This two-step verification is critical as it underlines the function's reversible nature.
Verification isn't just about plugging values in and carrying out operations. It's a deeper mathematical assurance that the relationships between functions carry resilience and consistency.

Function Operations

In mathematics, function operations refer to the different ways in which two or more functions can interact. These operations include addition, subtraction, multiplication, division, and composition, as seen in this exercise. The operations culminate in forming new functions or relationships.
In our scenario, the primary operation is the composition. With the two given functions, \( f(x) \) and \( g(x) \), the operation involves using each function within the other. This composition isn't just valid in itself; it's also instrumental in establishing inverses.

• Composition is used to check the inverse relationship.
• Returning the original input after operations (as seen with \( f(g(x)) = x \) and \( g(f(x)) = x \)) cements this understanding.

Function operations, therefore, form the bedrock of understanding how different algebraic expressions interact, setting the stage for more complex mathematical problem-solving.