Chapter 1: Problem 39
Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=(x-1)^{3} $$
Short Answer
Expert verified
The inverse function is \( f^{-1}(x) = x^{1/3} + 1 \).
Step by step solution
01
Define the function and set up the inverse
The given function is \( f(x) = (x - 1)^3 \). To find the inverse, we start by setting \( y = f(x) \), so \( y = (x - 1)^3 \). The goal of finding the inverse function \( f^{-1}(x) \) is to express \( x \) in terms of \( y \).
02
Solve for x in terms of y
Starting from \( y = (x - 1)^3 \), we solve for \( x \). First, take the cube root of both sides to obtain \( y^{1/3} = x - 1 \). Add 1 to both sides to solve for \( x \): \( x = y^{1/3} + 1 \). Thus, the inverse function is \( f^{-1}(x) = x^{1/3} + 1 \).
03
Verify that \(f^{-1}(f(x))=x\)
Substitute \( x \) into the inverse function: \( f^{-1}(f(x)) = f^{-1}((x - 1)^3) \). Now, replace within the inverse, \( f^{-1}(t) = t^{1/3} + 1 \), where \( t = (x - 1)^3 \). Applying the inverse, \( (x - 1)^3 \) under the cube root is simply \( x - 1 \). Adding 1 gives us \( x - 1 + 1 = x \). Thus, \( f^{-1}(f(x)) = x \).
04
Verify that \(f(f^{-1}(x))=x\)
Substitute the inverse into the original function: \( f(f^{-1}(x)) = f(x^{1/3} + 1) \). This gives \( ((x^{1/3} + 1) - 1)^3 \). Simplify inside the parenthesis to \( x^{1/3} \) and then cube it to get back to \( x \). Therefore, \( f(f^{-1}(x)) = x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Verifying Inverse Functions
The concept of verifying inverse functions is essential in understanding how two functions relate to each other as reverses. When verifying if two functions are truly inverses, you need to check two conditions:
The reason this works relies on the basic property of inverse functions: they "undo" each other's operation. For any input \( x \), applying a function and then its inverse function should return the original \( x \). In our example, we verified:
- The composition of the function and its inverse, \( f(f^{-1}(x)) \), should equal \( x \).
- Similarly, the composition of the inverse and the function, \( f^{-1}(f(x)) \), should also equal \( x \).
The reason this works relies on the basic property of inverse functions: they "undo" each other's operation. For any input \( x \), applying a function and then its inverse function should return the original \( x \). In our example, we verified:
- For \( f(f^{-1}(x)) = x \), substituting \( f^{-1} \) into \( f \) must give back the original \( x \).
- For \( f^{-1}(f(x)) = x \), applying the inverse to the function should also revert to \( x \).
Algebraic Manipulation
Algebraic manipulation is the backbone of solving equations, especially when dealing with inverse functions. It allows us to rearrange formulas to uncover hidden relationships between variables. In the context of inverse functions, it helps us isolate and understand the expression in its inverse form.
Let's look specifically at how algebraic manipulation was used in this problem:
Let's look specifically at how algebraic manipulation was used in this problem:
- Starting with \( y = (x - 1)^3 \), we needed to solve for \( x \), which involved reversing the operations applied to \( x \).
- We first removed the cube by taking the cube root, resulting in \( y^{1/3} = x - 1 \).
- Algebraic manipulation continued as we moved the terms around to solve for \( x \), adding 1 to both sides to isolate \( x \), giving \( x = y^{1/3} + 1 \).
Finding Inverse Formulas
Finding inverse formulas is not just about solving equations. It's the art of reversing a function's operations. The process of finding an inverse function essentially requires swapping input and output and solving for the new output. In our context, with the function \( f(x) = (x - 1)^3 \), we sought \( f^{-1}(x) \).
The primary steps involved are:
The primary steps involved are:
- Swap \( f(x) \) with \( y \), and write \( y = (x - 1)^3 \).
- Then solve for \( x \), turning each operation inside out – namely, we take the cube root, reversing the cube operation.
- This yields \( x = y^{1/3} + 1 \) which is \( f^{-1}(x) = x^{1/3} + 1 \).