Problem 19 - Intermediate Algebra

Question

Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=3 x\) and \(g(x)=\frac{1}{3} x\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = x\) and \((g \circ f)(x) = x\); both functions are inverses.

Step 1: Understand Function Composition

Function composition is when you apply one function to the results of another. Denoted by \((f \circ g)(x)\), it means apply \(g\) to \(x\), and then \(f\) to the result. Similarly, \((g \circ f)(x)\) means first apply \(f\) to \(x\), then \(g\) to that result.

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

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Since \(g(x) = \frac{1}{3}x\), we have \((f \circ g)(x) = f(g(x)) = f\left(\frac{1}{3}x\right) = 3\left(\frac{1}{3}x\right) = x\).

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

Now, compute \((g \circ f)(x)\) by substituting \(f(x)\) into \(g(x)\). Since \(f(x) = 3x\), we have \((g \circ f)(x) = g(f(x)) = g(3x) = \frac{1}{3}(3x) = x\).

Step 4: Verify Inverse Relationship

The calculations show that both \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\). This confirms that \(f\) and \(g\) are inverse functions of each other, as applying one after the other returns the original input.

Key Concepts

Inverse Functions

Inverse functions are pairs of functions that essentially 'undo' each other. For any function \(f\), there may exist another function \(g\) such that applying \(f\) first and then \(g\), or vice versa, brings you back to your starting point. This means that \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\) for all \(x\) in the domain of \(f\) and \(g\).

For example, in our exercise, \(f(x) = 3x\) and \(g(x) = \frac{1}{3}x\). Here, \(f\) multiplies its input by 3, and \(g\) does the reverse by dividing the input by 3. When you compose these functions, they effectively cancel each other out:

• \(f(g(x)) = f\left(\frac{1}{3}x\right) = 3\left(\frac{1}{3}x\right) = x\)
• \(g(f(x)) = g(3x) = \frac{1}{3}(3x) = x\)

These calculations indicate that \(f\) and \(g\) are inverses, always leading back to the original value \(x\).

Understanding inverse functions is a crucial foundation for solving equations and understanding many mathematical concepts in algebra and calculus.

Algebra

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and expressions. This makes it easier to handle general mathematical situations and solve equations.

When dealing with functions like \(f(x) = 3x\) and \(g(x) = \frac{1}{3}x\), algebra helps us to understand their behavior and relationships. Specifically, for two functions to be inverses, their compositions \((f \circ g)\) and \((g \circ f)\) need to return the identity function, \(x\).

Through algebraic manipulation, you can find:

• \((f \circ g)(x) = f(g(x)) = 3\left(\frac{1}{3}x\right) = x\)
• \((g \circ f)(x) = g(f(x)) = \frac{1}{3}(3x) = x\)

These results show that the identities are satisfied, confirming that the two functions are indeed inverses.

Algebra not only helps us to verify the inversion through composition but also simplifies expressions, aids in problem-solving, and remains a core element of higher mathematics.

Mathematical Proof

A mathematical proof is a logical argument that establishes the truth of a mathematical statement. In our exercise, the objective was to verify that two functions \(f(x) = 3x\) and \(g(x) = \frac{1}{3}x\) are inverse functions. This was done by proving that \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\).

To prove these statements, we used function composition and algebraic manipulation:

• We first substituted \(g(x)\) into \(f(x)\) to check if \((f \circ g)(x)\) equals \(x\), i.e., \(3\left(\frac{1}{3}x\right) = x\).
• We then substituted \(f(x)\) into \(g(x)\) to check if \((g \circ f)(x)\) equals \(x\), i.e., \(\frac{1}{3}(3x) = x\).

These steps clearly show that each function undoes the effect of the other, providing a solid proof of their inverse relationship.

Proofs are essential in mathematics to ensure that statements hold universally, regardless of initial assumptions or intuitive conclusions. It instills confidence that certain properties and relationships are valid under the defined conditions.