Problem 202 - Calculus Volume 1

Question

Use composition to determine which pairs of functions are inverses. \(f(x)=\frac{2}{3} x+2, g(x)=\frac{3}{2} x+3\)

Step-by-Step Solution

Answer

The functions are not inverses.

Step 1: Find the Composition f(g(x))

To check if two functions are inverses, compose them. Start with finding \(f(g(x))\). Substitute \(g(x)\) into \(f(x)\): \[f(g(x)) = f\left(\frac{3}{2}x + 3\right) = \frac{2}{3}\left(\frac{3}{2}x + 3\right) + 2\].

Step 2: Simplify f(g(x))

Distribute \(\frac{2}{3}\) into the expression inside the parenthesis: \[f\left(\frac{3}{2}x + 3\right) = \left(\frac{2}{3} \times \frac{3}{2}\right)x + \left(\frac{2}{3} \times 3\right) + 2 \]. Simplifying inside gives \(x + 2 + 2\). So, \[f(g(x)) = x + 4\].

Step 3: Find the Composition g(f(x))

Next, find \(g(f(x))\). Substitute \(f(x)\) into \(g(x)\): \[g(f(x)) = g\left(\frac{2}{3}x + 2\right) = \frac{3}{2}\left(\frac{2}{3}x + 2\right) + 3\].

Step 4: Simplify g(f(x))

Distribute \(\frac{3}{2}\) into the expression: \[g\left(\frac{2}{3}x + 2\right) = \left(\frac{3}{2} \times \frac{2}{3}\right)x + \left(\frac{3}{2} \times 2\right) + 3 \]. Simplifying gives \(x + 3 + 3\). So, \[g(f(x)) = x + 6\].

Step 5: Conclusion: Analyze Composition Results

For two functions to be inverses, both compositions \(f(g(x))\) and \(g(f(x))\) must return \(x\). Here, \(f(g(x)) = x + 4\) and \(g(f(x)) = x + 6\). Thus, \(f(x)\) and \(g(x)\) are not inverses.

Key Concepts

Composition of Functions

To determine if two functions are inverses, one handy tool we use is the composition of functions. This involves nesting one function inside another. Suppose we have two functions, say \(f(x)\) and \(g(x)\). We then look at the result of applying one function to the result of the other. In algebra, this is written as \(f(g(x))\) or \(g(f(x))\).

- **Nested Operations:** Begin by plugging \(g(x)\) directly into the expression for \(f(x)\) if you're composing \(f(g(x))\), and vice versa for \(g(f(x))\).
- **Chain Together:** Essentially, you are "chaining" the functions in a way that sees one function picking up where the other leaves off. In our example, \(f(g(x))\) means taking the output of \(g(x)\) and using it as the input for \(f(x)\).

This technique helps in verifying if two functions \(f(x)\) and \(g(x)\) are true inverses of each other. If the composition results in just the variable \(x\) itself, they are inverses.

Algebraic Manipulation

Once you set up the function compositions, the next crucial skill is algebraic manipulation, which involves simplifying the expressions. When dealing with function compositions like \(f(g(x))\) or \(g(f(x))\), simplifying them gives insights into the relationships between these functions.

- **Distribution:** Often, simplifying involves distribution, as seen in our example: \(f(g(x)) = \frac{2}{3}\left(\frac{3}{2}x + 3\right) + 2\). Here, distribute the multiplier \(\frac{2}{3}\) across each term within the parenthesis.
- **Combine Like Terms:** Afterwards, merge similar terms. In the scenario, we've distributed \(\frac{2}{3}\) and then grouped terms: resulting in \(x + 4\).

Simplifying such complex expressions using algebraic manipulation clarifies whether specific compositions simplify down to \(x\). This step is vital in determining if functions cancel each other in a way that marks them as true inverses.

Function Properties

Understanding the properties of functions is the bedrock of determining if two functions are inverses. The main property of interest here is that inverse functions will "undo" each other perfectly.

- **Inverse Characteristic:** When \(f\) and \(g\) are inverses, applying one to the result of the other yields the starting value \(x\). Both \(f(g(x)) = x\) and \(g(f(x)) = x\) should hold true.
- **Checking Outcomes:** We found \(f(g(x)) = x + 4\) and \(g(f(x)) = x + 6\) in our check. Since neither simplified precisely to \(x\), this shows that \(f(x)\) and \(g(x)\) aren’t inverses.

Understanding these function properties allows us to conclude on their inverse status. By meeting the criterion of simplification to \(x\), functions can be verified as rightful inverses to one another.