Problem 25 - Thomas Calculus

Question

Each of Exercises \(25-34\) gives a formula for a function \(y=f(x)\) . In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1}\) . As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x.\) $$f(x)=x^{5}$$

Step-by-Step Solution

Answer

\(f^{-1}(x) = x^{1/5}\), domain and range are all real numbers. Verified by substituting.

Step 1: Understand the Function

We are given the function \(f(x) = x^5\), which is a simple power function. Our task is to find the inverse of this function, \(f^{-1}(x)\).

Step 2: Swap and Solve

To find the inverse of \(f(x)\), we start by replacing \(y\) for \(f(x)\): \(y = x^5\). Next, we swap \(x\) and \(y\) to get \(x = y^5\) and solve for \(y\). This gives us the equation \(y = x^{1/5}\). Therefore, \(f^{-1}(x) = x^{1/5}\).

Step 3: Identify Domain and Range of Inverse Function

The domain of the inverse function \(f^{-1}(x)\) is all real numbers because the fifth root of any real number is also a real number. The range of \(f^{-1}\) is also all real numbers since it can produce any real number output.

Step 4: Verify Inverse Functions

To verify that \(f^{-1}(x)\) is correct, we need to show that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). Calculate: \(f(f^{-1}(x)) = f(x^{1/5}) = (x^{1/5})^5 = x\). Also calculate: \(f^{-1}(f(x)) = f^{-1}(x^5) = (x^5)^{1/5} = x\). Both expressions simplify to \(x\), verifying correctness.

Key Concepts

Domain and Range

When we talk about the domain of a function, we are referring to all the possible input values the function can accept. For example, in the case of the inverse function of a power function like \(f(x) = x^5\), the inverse function is \(f^{-1}(x) = x^{1/5}\). Let's look at what this means for domain and range.

1\. **Domain of \(f^{-1}(x)\):** The domain refers to the set of all possible values of \(x\) that the function can take. For \(f^{-1}(x) = x^{1/5}\), any real number can be input, therefore, its domain is all real numbers. The fifth root of any real number is still a real number.

2\. **Range of \(f^{-1}(x)\):** When we consider the range, we are thinking about all the possible output values of the function. Since \(f^{-1}(x) = x^{1/5}\) can produce any real number, the range is also all real numbers. Any real number output can result from taking the fifth root of another real number. This highlights that the domain and range for this particular inverse function are both unrestricted across the entire real number line.

Power Function

Power functions are very straightforward: they are functions in the form \(f(x) = x^n\), where \(n\) is a real number. In our original problem, the power function is \(f(x) = x^5\). These functions can have different characteristics depending on the power, \(n\), so let's explore their properties.

• **Degree:** The degree of a power function indicates the exponent. Here it's \(5\), which means we're working with a quintic function.
• **Monotonic Behavior:** For odd powers like \(5\), power functions are monotonic, meaning they are either always increasing or always decreasing. In our example, \(f(x) = x^5\) is always increasing.
• **Symmetry:** Odd power functions like \(x^5\) are symmetric around the origin. This means that \(f(-x) = -f(x)\).

Understanding such properties helps with finding inverses because the increasing behavior ensures a one-to-one nature, which is necessary for an inverse to exist.

Function Inverse Verification

Verifying the correctness of an inverse function is crucial, because it ensures that the operations have reversed the original function perfectly. In the exercise, the original function was \(f(x) = x^5\) and its claimed inverse is \(f^{-1}(x) = x^{1/5}\). There are two main ways to verify this relationship:

1\. **Forward Verification: \(f(f^{-1}(x)) = x\)**
Replace \(f^{-1}(x)\) in \(f(x)\):
\[f(x^{1/5}) = (x^{1/5})^5 = x\]
This shows that applying \(f\) to its inverse recovers the input, \(x\).

2\. **Backward Verification: \(f^{-1}(f(x)) = x\)**
Replace \(f(x)\) in \(f^{-1}(x)\):
\[f^{-1}(x^5) = (x^5)^{1/5} = x\]
This verifies that the inverse function also undoes \(f(x)\) when computed backward.

Both verifications affirm that \(f^{-1}(x)\) is indeed the correct inverse of \(f(x)\). This ensures a complete and reliable understanding of how the inverse function accurately reflects the original function's values.