Problem 202 - Calculus Volume 1

Question

For the following exercises, 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: Understanding the Problem

We need to determine if the two functions \( f(x) = \frac{2}{3} x + 2 \) and \( g(x) = \frac{3}{2} x + 3 \) are inverses of each other. To do that, we have to check if both \( f(g(x)) = x \) and \( g(f(x)) = x \); in other words, when we compose each function with the other, we should end up with \( x \).

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

To compute \( f(g(x)) \), substitute \( g(x) = \frac{3}{2} x + 3 \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{3}{2} x + 3\right) = \frac{2}{3}\left(\frac{3}{2} x + 3\right) + 2 \]Simplify: \[ = \left(1\times x + \frac{2}{3}\times 3\right) + 2 = x + 2 + 2 = x + 4 \]Since \( f(g(x)) eq x \), this means \( g(x) \) is not the inverse of \( f(x) \).

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

Now, compute \( g(f(x)) \) by substituting \( f(x) = \frac{2}{3} x + 2 \) into \( g(x) \): \[ g(f(x)) = g\left(\frac{2}{3} x + 2\right) = \frac{3}{2}\left(\frac{2}{3} x + 2\right) + 3 \]Simplify: \[ = \left(x + \frac{3}{2}\times 2\right) + 3 = x + 3 + 3 = x + 6 \]Since \( g(f(x)) eq x \), this also confirms that \( f(x) \) is not the inverse of \( g(x) \).

Step 4: Conclusion

Since neither \( f(g(x)) = x \) nor \( g(f(x)) = x \), the functions \( f(x) \) and \( g(x) \) are not inverses of each other.

Key Concepts

Function Composition

Function composition is when you apply one function to the results of another function. This means, you take the output of one function and use it as the input for another. For example, when finding if functions are inverses, you use composition to check results.
In our exercise, we check by composing two functions, \( f(x) = \frac{2}{3} x + 2 \) and \( g(x) = \frac{3}{2} x + 3 \).
You need to see if when you plug \( g(x) \) into \( f(x) \), the outcome is the original \( x \), and vice-versa.

• To compose \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \).
• If the result is just \( x \), then that means \( g(x) \) might be an inverse of \( f(x) \).
• Do the same for \( g(f(x)) \). If it's \( x \), then \( f(x) \) might be an inverse of \( g(x) \).

In our problem, neither \( f(g(x)) = x \) nor \( g(f(x)) = x \) hold true, showing the functions are not inverses, using composition.

Function Inverses

The concept of function inverses involves finding another function that reverses the effect of the original one. Essentially, if \( f(x) \) and \( g(x) \) are inverses, applying one function and then the other should bring you back to where you started with \( x \).
To be precise, inverses follow these conditions:

• \( f(g(x)) = x \)
• \( g(f(x)) = x \)

These equations mean if you apply \( g(x) \) first and then \( f(x) \), you should get \( x \) back, and vice versa.
In mathematical terms, if these conditions do not hold true, as shown in our exercise, then the functions are not inverses. Using this test is central to determining inverses. If you find any other result than \( x \), like \( x+4 \) or \( x+6 \) as in our example, the functions definitely are not inverses of each other.

Algebraic Manipulation

Algebraic manipulation is crucial in solving and simplifying expressions, especially during composition. It involves playing around with algebraic expressions to reveal deeper relationships or check conditions.
During the exercise, we applied algebraic manipulation to simplify function compositions, such as solving \( f(g(x)) \) and \( g(f(x)) \).
Here's how it helps:

• Identify and simplify shared factors or values. For example, recognize constant terms that affect outcomes.
• Using distribution: when we had \( f(g(x)) = \frac{2}{3}\left(\frac{3}{2} x + 3\right) + 2 \), distribute \( \frac{2}{3} \) properly to solve.
• Simplify expressions to find if they exactly equal \( x \) or not. In our example, this led to discovering they did not equal \( x \).

Such steps can uncover relationships or lack thereof between functions. This manipulation often decides outcomes regarding whether functions are inverses.