Problem 34 - Intermediate Algebra

Question

Find the inverse of the given function by using the process illustrated in Examples 3 and 4 of this section, and then verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). $$f(x)=-6 x+2$$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = -\frac{x}{6} + \frac{1}{3}\), and it satisfies the conditions \((f \circ f^{-1})(x) = x\) and \((f^{-1} \circ f)(x) = x\).

Step 1: Replace f(x) with y

Begin by replacing the function notation \(f(x)\) with \(y\) to make it easier to manipulate. So, \( y = -6x + 2 \).

Step 2: Swap x and y

Interchange the variables \(x\) and \(y\) to find the inverse function. This gives you \( x = -6y + 2 \).

Step 3: Solve for y

Isolate \(y\) to solve for it in terms of \(x\). Start by subtracting 2 from both sides to get \( x - 2 = -6y \). Then, divide by -6: \( y = \frac{x - 2}{-6} \). Simplify it further to get \( y = -\frac{x}{6} + \frac{1}{3} \).

Step 4: Substitute back to function notation

Rewrite the expression for \(y\) using inverse function notation. The inverse function is \( f^{-1}(x) = -\frac{x}{6} + \frac{1}{3} \).

Step 5: Verify \((f \circ f^{-1})(x) = x\)

Verify by substituting \(f^{-1}(x)\) into \(f(x)\). Compute \(f(f^{-1}(x)) = -6(-\frac{x}{6} + \frac{1}{3}) + 2 \). This simplifies to \( x \), thus showing \( (f \circ f^{-1})(x) = x \).

Step 6: Verify \((f^{-1} \circ f)(x) = x\)

Verify by substituting \(f(x)\) into \(f^{-1}(x)\). Compute \(f^{-1}(f(x)) = -\frac{-6x + 2}{6} + \frac{1}{3} \). This simplifies to \( x \), thus showing \( (f^{-1} \circ f)(x) = x \).

Key Concepts

Function Composition

Function composition involves combining two functions such that the output of one becomes the input of the other. Imagine it as two gears working together where the spinning of one gear causes the other to turn. For example, if you have a function \( f(x) \) and another function \( g(x) \), their composition \( (f \circ g)(x) \) yields another function where \( f \) is applied to the result of \( g(x) \).
In this context, we are working with \( (f \circ f^{-1})(x) \) and \( (f^{-1} \circ f)(x) \). Here, \( f(x) \) and \( f^{-1}(x) \) are composed to yield \( x \), effectively reversing the operations of each function. This demonstrates a key property of inverse functions: composing a function with its inverse yields the original input.
This back-and-forth motion exemplifies why the idea of inverses is so pivotal in mathematics.

Inverse Verification

Inverse verification is the process of confirming that two functions \( f(x) \) and \( f^{-1}(x) \) are indeed inverses of each other. This is shown by verifying \((f \circ f^{-1})(x) = x\) and \((f^{-1} \circ f)(x) = x\). Think of it as a double-check: ensuring that both directions of operation return the starting value.
To verify that \( f(x) = -6x + 2 \) and its inverse \( f^{-1}(x) = -\frac{x}{6} + \frac{1}{3} \) are correct, they must fulfill both conditions mentioned. Each essentially involves substituting each function into the other rigorously. When correctly substituted, the operations simplify back to \( x \), guaranteeing that the transformations are perfectly valid.

Function Manipulation

Function manipulation involves altering the form or expressions of functions, allowing us to solve for and work with different variables. In finding inverses, we manipulate the function’s equation to switch the roles of \( x \) and \( y \).
Initially, the function \( y = -6x + 2 \) is restructured by swapping \( x \) and \( y \), resulting in \( x = -6y + 2 \). Solving for \( y \) means manipulating the equation by performing operations like addition, subtraction, multiplication, or division. Here, we subtract 2 from each side and then divide by -6 to isolate \( y \).
Such calculations are the cornerstone of solving algebraic equations within function manipulation. The result, \( y = -\frac{x}{6} + \frac{1}{3} \), offers a new perspective on the same relationship, now seen through its inverse.

Algebraic Equations

Algebraic equations form the bedrock of much of mathematics, serving as statements that express equality between two expressions. Finding the inverse of a function, for instance, is an exercise in working through algebraic equations to rearrange and isolate variables.
Solving these equations often requires a step-by-step approach where terms are moved systematically to isolate the desired variable. When finding the inverse of \( f(x) = -6x + 2 \), the goal is to isolate \( y \) through a series of steps: first by subtracting 2, then dividing by -6. Each manipulation ensures that both sides of the equation remain equal.
Algebraic equations are all about balance and order, and through precise manipulation, we unveil the inverse function \( f^{-1}(x) \). This demystifies the sequential operations required to shift equations from one form to another. By mastering these steps, students can handle a wide range of algebraic challenges.