Problem 23 - Intermediate Algebra

Question

Find the inverse of the given function by using the "undoing process," and then verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). (Objective 4) $$f(x)=-2 x+1$$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = -\frac{x}{2} + \frac{1}{2} \). Verification shows \( (f \circ f^{-1})(x) = x \) and \( (f^{-1} \circ f)(x) = x \).

Step 1: Identify the Function

Our given function is \( f(x) = -2x + 1 \). Our goal is to find its inverse function \( f^{-1}(x) \).

Step 2: Switch Roles of x and y

To find the inverse, we first replace \( f(x) \) with \( y \), so the equation is \( y = -2x + 1 \). Next, switch the roles of \( x \) and \( y \) to get \( x = -2y + 1 \).

Step 3: Solve for y

Now, solve the equation \( x = -2y + 1 \) for \( y \). Subtract 1 from both sides: \( x - 1 = -2y \). Then, divide by -2: \( y = \frac{x - 1}{-2} \), or \( y = -\frac{x}{2} + \frac{1}{2} \). This is our \( f^{-1}(x) \).

Step 4: Verify the Inverse with f(f^-1)

To verify, we need to show that \( (f \circ f^{-1})(x) = x \). Substitute \( f^{-1}(x) = -\frac{x}{2} + \frac{1}{2} \) into \( f(x) \): \( f(f^{-1}(x)) = f(-\frac{x}{2} + \frac{1}{2}) = -2(-\frac{x}{2} + \frac{1}{2}) + 1 \). Simplifying gives \( x \), confirming this part.

Step 5: Verify the Inverse with f^-1(f)

Now verify that \( (f^{-1} \circ f)(x) = x \). Substitute \( f(x) = -2x + 1 \) into \( f^{-1}(x) \): \( f^{-1}(f(x)) = f^{-1}(-2x + 1) = -\frac{-2x + 1}{2} + \frac{1}{2} \). Simplify this expression to get \( x \), confirming this part.

Key Concepts

Undoing Process

The undoing process is a method used to find the inverse of a function. Think of it as reversing the steps needed to apply the function. If a function describes a certain process or operation to get from input to output, the undoing process reverses those steps to retrieve the original input from the output.
For example, in the case of the function \(f(x) = -2x + 1\), the operations are multiplying by -2 and then adding 1. To "undo" this, you reverse the steps:

• First, replace the function notation \(f(x)\) with \(y\), giving \(y = -2x + 1\).
• Switch the roles of \(x\) and \(y\) to represent the inverse relationship, resulting in \(x = -2y + 1\).
• Solve this equation for \(y\) to find \(y = f^{-1}(x)\).

This new equation, rearranged back to get \(y\) alone, represents the undoing of the original function's steps.

Function Verification

Once you find the inverse of a function, it's crucial to verify it to ensure accuracy. Verification involves checking that applying the original function and then the inverse gets you back to your starting point. You need to prove that both \((f \circ f^{-1})(x) = x\) and \((f^{-1} \circ f)(x) = x\) hold true.
For our function \(f(x) = -2x + 1\) and its inverse \(f^{-1}(x) = -\frac{x}{2} + \frac{1}{2}\), this means:

• Verify \((f \circ f^{-1})(x)\): Substituting the inverse into the function gives \(f(f^{-1}(x)) = -2(-\frac{x}{2} + \frac{1}{2}) + 1\). Simplify this, and you'll find it equals \(x\).
• Verify \((f^{-1} \circ f)(x)\): Substitute the function into its inverse, giving \(f^{-1}(-2x + 1) = -\frac{-2x + 1}{2} + \frac{1}{2}\). Simplifying this also gets you back to \(x\).

Both verifications confirm that the inverse is correct.

Intermediate Algebra

Understanding inverse functions is a crucial part of intermediate algebra. This concept consolidates various algebraic techniques, such as solving equations and working with different function forms. When dealing with inverses, algebra is your tool to methodically switch and solve expressions.
Consider the step of solving \(x = -2y + 1\) for \(y\):

• Subtract 1 from both sides to isolate terms with \(y\): \(x - 1 = -2y\).
• Divide by -2 to isolate \(y\): \(y = \frac{x - 1}{-2}\) or equivalently \(y = -\frac{x}{2} + \frac{1}{2}\).

These steps are an example of intermediate algebra in action. Being comfortable with these manipulations is essential for correctly finding function inverses and verifying them effectively. Remember, practice makes perfect in mastering the skills needed for more advanced mathematical concepts.