Problem 25 - Intermediate Algebra

Question

Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=-\frac{1}{4} x-\frac{1}{2}\) and \(g(x)=-4 x-2\)

Step-by-Step Solution

Answer

Both compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) simplify to \(x\).

Step 1: Define the Composition (f ◦ g)(x)

The composition \((f \circ g)(x)\) means that you substitute the function \(g(x)\) into the function \(f(x)\). This results in \(f(g(x))\). First, evaluate \(g(x)\) given \(x\).

Step 2: Substitute g(x) Into f(x)

Since \(g(x) = -4x - 2\), substitute \(-4x - 2\) into \(f(x) = -\frac{1}{4}x - \frac{1}{2}\).\[f(g(x)) = f(-4x - 2) = -\frac{1}{4}(-4x - 2) - \frac{1}{2}\]

Step 3: Simplify the Expression for (f ◦ g)(x)

Simplify inside the function:\[-\frac{1}{4}(-4x - 2) = x + \frac{1}{2}\]Combine with the remaining terms:\[x + \frac{1}{2} - \frac{1}{2} = x\]

Step 4: Define the Composition (g ◦ f)(x)

The composition \((g \circ f)(x)\) means that you substitute the function \(f(x)\) into the function \(g(x)\). This results in \(g(f(x))\). First, evaluate \(f(x)\) given \(x\).

Step 5: Substitute f(x) Into g(x)

Since \(f(x) = -\frac{1}{4}x - \frac{1}{2}\), substitute \(-\frac{1}{4}x - \frac{1}{2}\) into \(g(x) = -4x - 2\).\[g(f(x)) = g\left(-\frac{1}{4}x - \frac{1}{2}\right) = -4\left(-\frac{1}{4}x - \frac{1}{2}\right) - 2\]

Step 6: Simplify the Expression for (g ◦ f)(x)

Simplify inside the function:\[-4\left(-\frac{1}{4}x - \frac{1}{2}\right) = x + 2 \]Combine with the remaining terms:\[x + 2 - 2 = x\]

Step 7: Conclusion

Both compositions result in the identity function:\((f \circ g)(x) = x\) and \((g \circ f)(x) = x\).

Key Concepts

Inverse Functions

Inverse functions are essential in mathematics as they allow us to "undo" the operations of a given function. Imagine having two functions, such as the ones in the original problem: \(f(x) = -\frac{1}{4}x - \frac{1}{2}\) and \(g(x) = -4x - 2\). Inverse functions are pairs of functions \(f\) and \(g\) such that when one function operates on the output of the other, the original input is recovered. This is represented in function notation as:

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

For instance, substituting \(g(x)\) into \(f(x)\), or vice versa, should ideally result in the identity function, which means we get back our original input \(x\). In simple terms, if \(f\) can undo the work of \(g\), then \(g\) can undo the work of \(f\), proving they are indeed inverses of each other.
Knowing how to identify and verify inverse functions is particularly useful in algebra, calculus, and beyond, as it allows us to solve equations and understand function behaviors deeply.

Algebraic Functions

Algebraic functions are a fundamental concept in algebra. These functions are built using algebraic expressions that involve a finite combination of numbers, variables, and operations such as addition, subtraction, multiplication, division, and taking roots.
The functions given in the problem, \(f(x) = -\frac{1}{4}x - \frac{1}{2}\) and \(g(x) = -4x - 2\), are linear algebraic functions since they are expressed as polynomials of degree 1.

• In \(f(x)\), the function is defined by scaling the input \(x\) by \(-\frac{1}{4}\) and then shifting it \(-\frac{1}{2}\).
• In \(g(x)\), the function scales the input \(x\) by \(-4\) and shifts it by \(-2\).

These expressions use fundamental algebraic operations to take an input \(x\) and map it to an output. When performing function composition or examining inverse relations, these simple linear transformation rules make the behavior of the functions straightforward to analyze and predict. Understanding these functions makes it easier to approach more complex expressions and operations.

Identity Function

The identity function is one of the simplest yet most significant types of functions in mathematics. Noted for its role as a 'do-nothing' function, it maps any input directly to itself. In mathematical terms, this is expressed as \(I(x) = x\) for all real numbers \(x\).
In the context of function composition, as shown in the problem, presenting both \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\) effectively signifies that the compositions of these pairs of functions result in the identity function. This means that no matter what value of \(x\) you start with, after applying these compositions, the result is simply \(x\) again.

• This property is directly tied to the concept of inverse functions, as the composition of a function and its inverse should yield the identity function.

The identity function is crucial because it helps us verify when two functions are truly inverses. Whenever their composition yields the identity function, we can confirm that they undo each other's effects perfectly, revealing a beautiful symmetry in mathematical operations.