Problem 70 - A Graphical Approach to College Algebra

Question

For certain pairs of functions \(f\) and \(g,(f \circ g)(x)=x\) and \((g \circ f)(x)=x .\) Show that this is true for the pairs. $$f(x)=-3 x, g(x)=-\frac{1}{3} x$$

Step-by-Step Solution

Answer

The functions are inverses: \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\).

Step 1: Understanding the Composition of Functions

First, recall the definition of function composition. The composition of two functions, \((f \circ g)(x)\), means you apply the function \(g\) first and then apply \(f\) to the result. Similarly, \((g \circ f)(x)\) indicates applying \(f\) first and then \(g\). To solve the exercise, compute both \(f(g(x))\) and \(g(f(x))\) and verify if in both cases the result is \(x\).

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

Begin by determining \(g(x)\): \[g(x) = -\frac{1}{3}x\] Next, substitute \(g(x)\) into \(f(x)\) to get \(f(g(x))\): \[f(g(x)) = f\left(-\frac{1}{3}x\right) = -3\left(-\frac{1}{3}x\right)\] Calculate the expression: \[-3\times -\frac{1}{3}x = x\] Thus, \((f \circ g)(x) = x\).

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

Now determine \(f(x)\): \[f(x) = -3x\] Substitute \(f(x)\) into \(g(x)\) to get \(g(f(x))\): \[g(f(x)) = g(-3x) = -\frac{1}{3}(-3x)\] Calculate the expression: \[-\frac{1}{3}\times -3x = x\] Thus, \((g \circ f)(x) = x\).

Step 4: Conclude with Verification

We have shown that both function compositions result in \(x\). Specifically, \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\), verifying the desired property of these functions. Thus, \(f\) and \(g\) are inverses of each other.

Key Concepts

Composition of Functions

Composition of functions is a fundamental concept in mathematics. It involves applying one function to the results of another. Think of it like a recipe: first, you follow one set of instructions, then the next. When we write

• \((f \circ g)(x)\), it means that you first calculate \(g(x)\), and then take that result as the input for \(f(x)\).
• Similarly, for \((g \circ f)(x)\), you first find \(f(x)\), and then use that as the input for \(g(x)\).

To compute these compositions:

• Start with the inner function. For \((f \circ g)(x)\), compute \(g(x) = -\frac{1}{3}x\).
• Then apply the outer function, \(f\), to this result: \(f\left(-\frac{1}{3}x\right) = -3\left(-\frac{1}{3}x\right) = x\).

In this way, understanding how functions compose is rather straightforward when tackled step-by-step. Integrating functions involves understanding their order and substitution—just as solving a puzzle one piece at a time.

Algebraic Proof

Algebraic proof is used to verify or demonstrate certain mathematical properties. It's like logically proving that what you're solving is true. In the exercise, we are asked to prove two compositions that yield the identity function (returning the original input):

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

Firstly, we show \((f \circ g)(x)\) by computing step-by-step:

• Take \(g(x) = -\frac{1}{3}x\)
• Calculate \(f(g(x)) = -3(-\frac{1}{3}x) = x\)

Then, for \((g \circ f)(x)\):

• Find \(f(x) = -3x\)
• Compute \(g(f(x)) = -\frac{1}{3}(-3x) = x\)

In algebraic terms, showing these steps establishes that both function compositions result in \(x\), confirming that the algebraic operations are valid. This proof methodically demonstrates that these functions perform as expected. By confirming both compositions equal the identity function, we establish the inverses.

Inverse Functions

Inverse functions are special because they "undo" each other. For two functions to be inverses, applying them one after the other must return the original input. In the exercise, we've shown:

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

Which confirms that \(f\) and \(g\) are indeed inverses.
Think of inverse functions like wearing and then removing gloves. If \(f(x)\) is putting the gloves on, \(g(x)\) would be taking them off. After removing gloves, your hands are back to their original state.
To check if two functions are inverses:

• Show that both \((f \circ g)(x)\) and \((g \circ f)(x)\) return the input \(x\).
• This requirement ensures the first function's effects are nullified by the second.

Inverse functions play a crucial role in algebra, allowing one to determine pre-images and undo transformations systematically, as demonstrated fully by the functions \(f(x) = -3x\) and \(g(x) = -\frac{1}{3}x\). Such reversibility is a fascinating aspect of mathematical functions.