Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(31-44\), find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=x+1 $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = x - 1\); both composition checks confirm it is correct.

Step 1: Understanding the Inverse Function

The inverse function, denoted as \(f^{-1}(x)\), reverses the operation of the original function \(f(x)\). For \(f(x) = x + 1\), the inverse function \(f^{-1}(x)\) would subtract 1 to return to the original input value.

Step 2: Solve for the Inverse Function

To find \(f^{-1}(x)\), we start with the equation \(y = x + 1\) and solve for \(x\):\[y = x + 1\]Subtract 1 from both sides to isolate \(x\):\[x = y - 1\]Thus, \(f^{-1}(x) = x - 1\).

Step 3: Verify \(f^{-1}(f(x)) = x\)

Substitute \(f(x)\) into \(f^{-1}(x)\):\[f^{-1}(f(x)) = f^{-1}(x + 1)\]Since \(f^{-1}(x) = x - 1\),\[f^{-1}(x + 1) = (x + 1) - 1 = x\]This verifies that \(f^{-1}(f(x)) = x\).

Step 4: Verify \(f(f^{-1}(x)) = x\)

Substitute \(f^{-1}(x)\) into \(f(x)\):\[f(f^{-1}(x)) = f(x - 1)\]Since \(f(x) = x + 1\),\[f(x - 1) = (x - 1) + 1 = x\]This verifies that \(f\left(f^{-1}(x)\right) = x\).

Key Concepts

Function Verification

One important aspect of inverse functions is verifying them to ensure they truly are the inverses of the original functions. This involves checking two key conditions:

• The expression \( f^{-1}(f(x)) = x \).
• The expression \( f(f^{-1}(x)) = x \).

These conditions confirm that when you apply a function and its inverse successively, you return to your original input value.
To illustrate this with our example function \( f(x) = x + 1 \), we perform the following steps:

• First, compute \( f^{-1}(f(x)) \). After substituting \( f(x) \) into \( f^{-1}(x) \), you should end up with \( x \).
• Next, compute \( f(f^{-1}(x)) \). Substitute the inverse function back into the original function to verify that it returns \( x \).

In both cases, achieving the result \( x \) confirms the functions are correct inverses of each other and thoroughly verifies the function operations.

Inverse Operation

Inverse operations are foundational to understanding inverse functions. They essentially reverse the effect of a given function. Consider the function \( f(x) = x + 1 \). Here, the operation involves adding 1 to \( x \). Its inverse, \( f^{-1}(x) \), does precisely the opposite: it subtracts 1 from \( x \).
This reversing process is crucial because it enables us to "undo" the actions of a function. For every operation a function represents, the inverse performs a counter-operation.

• Addition in the function leads to subtraction in its inverse.
• Multiplication in the function results in division in the inverse.

For our example, the inverse operation is shown by solving \( y = x + 1 \) for \( x \). By rearranging to \( x = y - 1 \), we explicitly define the inverse operation as subtraction, leading us to \( f^{-1}(x) = x - 1 \). This simple model captures the essence of how inverses reverse function operations.

Algebraic Manipulation

Algebraic manipulation is the art of rearranging equations to isolate specific variables, particularly when finding inverse functions. When tasked with identifying an inverse, we typically start with the function's equation and solve it algebraically for the dependent variable.
In our function \( f(x) = x + 1 \), we represent it as \( y = x + 1 \). To find the inverse, perform steps to isolate \( x \):

• Step 1: Swap \( x \) and \( y \) as \( x = y + 1 \).
• Step 2: Subtract 1 from \( y \), leading to \( x = y - 1 \).

This process yields the inverse function \( f^{-1}(x) = x - 1 \). Such algebraic manipulations are powerful tools allowing us to reframe functions and unveil their inverses by changing their algebraic structure.
Through these manipulations, we not only derive the inverse but also develop a deeper understanding of the relationship between functions and their respective inverses.