Problem 27 - Algebra and Trigonometry

Question

Use the Inverse Function Property to show that \(f\) and \(g\) are inverses of each other. $$ f(x)=2 x-5 ; \quad g(x)=\frac{x+5}{2} $$

Step-by-Step Solution

Answer

\(f(x) = 2x - 5\) and \(g(x) = \frac{x+5}{2}\) are inverses because \(f(g(x)) = x\) and \(g(f(x)) = x\).

Step 1: Define the Inverse Function Property

Two functions, \(f\) and \(g\), are inverses of each other if for every \(x\) in the domain of \(f\), \(f(g(x)) = x\), and for every \(x\) in the domain of \(g\), \(g(f(x)) = x\). We will use this property to verify that \(f\) and \(g\) are inverses.

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

Start by calculating \(f(g(x))\). We know that \(g(x) = \frac{x+5}{2}\). Substitute \(g(x)\) into \(f(x) = 2x - 5\):\[f(g(x)) = f\left(\frac{x+5}{2}\right) = 2\left(\frac{x+5}{2}\right) - 5\]Simplify:\[= x+5 - 5 = x\]

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

Now calculate \(g(f(x))\). We have \(f(x) = 2x - 5\). Substitute \(f(x)\) into \(g(x) = \frac{x+5}{2}\):\[g(f(x)) = g(2x - 5) = \frac{(2x - 5) + 5}{2}\]Simplify:\[= \frac{2x}{2} = x\]

Step 4: Verify the Inverse Function Property

We found that \(f(g(x)) = x\) and \(g(f(x)) = x\) for all \(x\) in their respective domains. Therefore, according to the inverse function property, \(f\) and \(g\) are inverses of each other.

Key Concepts

Inverse Function Property

The inverse function property is a cornerstone in understanding inverse functions. This property states that if you have two functions, say \( f \) and \( g \), and they are inverses, then their compositions, \( f(g(x)) \) and \( g(f(x)) \), will both return the original input \( x \). This means that after applying one function, using the second function should reverse or "undo" the effect of the first one, taking you back to the starting value.
To verify if two functions are indeed inverses:

• Check if \( f(g(x)) = x \) for every \( x \) within the domain of \( g \).
• Check if \( g(f(x)) = x \) for every \( x \) within the domain of \( f \).

This double check ensures the two functions perfectly reverse each other's effect across the span of their usable values, also known as their domain. Without satisfying these conditions, the functions \( f \) and \( g \) cannot be labeled as true inverses.

Function Composition

Function composition is like applying one function after another. Imagine you have two functions, \( f \) and \( g \). The composition \( f(g(x)) \) means you're first giving \( x \) to \( g \), and whatever \( g(x) \) gives you, you then pass to \( f \).
In the problem, to check if \( f \) and \( g \) are inverses, composition is our tool. We found:

• First, that \( f(g(x)) = x \) by substituting \( g(x) = \frac{x+5}{2} \) into \( f \).
• Next, that \( g(f(x)) = x \) by substituting \( f(x) = 2x - 5 \) into \( g \).

Both results confirm that the composition of these functions always returns \( x \), demonstrating that \( f \) and \( g \) truly undo each other when composed in either order. Function composition thus acts as a verifiable pathway to determine if two functions are inverses.

Domain of Functions

The domain of a function refers to all the possible input values, usually denoted as \( x \), that the function can accept without running into undefined behavior, like division by zero. When dealing with inverse functions, understanding the domain becomes crucial because:

• The composition \( f(g(x)) \) should make sense for every \( x \) in the domain of \( g \).
• The composition \( g(f(x)) \) should make sense for every \( x \) in the domain of \( f \).

For example, with \( f(x) = 2x - 5 \) and \( g(x) = \frac{x+5}{2} \), there are no restrictions like division by zero or square roots of negatives, thus both functions operate efficiently over all real numbers, making their domains all real numbers. This unrestricted domain simplifies the verification process, allowing \( x \) to take any real value while checking the inverse property. Having a clear understanding of these domains ensures the functions operate correctly, providing a seamless experience without unexpected issues.