Problem 19
Question
Each of Exercises \(19-24\) 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
Verified Answer
The inverse function is \(f^{-1}(x) = x^{1/5}\), with domain \((-\infty, \\infty)\) and range \((-\infty, \\infty)\).
1Step 1: Understand the function
The given function is \( f(x) = x^5 \). This is a one-to-one function because each input corresponds to exactly one output. To find the inverse, we need to restrict the domain so that the function is bijective. Here, the domain is all real numbers \( \mathbb{R} \), and the range is also \( \mathbb{R} \).
2Step 2: Swap and solve for the inverse
To find \( f^{-1}(x) \), start by swapping \( x \) and \( y \), giving \( x = y^5 \). Solve this equation for \( y \) to get the inverse function. This involves taking the fifth root of both sides: \( y = \sqrt[5]{x} \). Thus, the inverse function is \( f^{-1}(x) = x^{1/5} \).
3Step 3: Determine the domain and range
Since the original function \(f(x) = x^5\) is defined for all real numbers and maps onto all real numbers, \(f^{-1}(x) = x^{1/5}\) will also be defined for all real numbers and map onto all real numbers. So, both the domain and the range of \(f^{-1}(x)\) are \(\mathbb{R}\).
4Step 4: Verify the inverse function
To confirm \(f^{-1}\) is correct, show \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). For the first, substitute \(f^{-1}(x)\) into \(f\): \(f(x^{1/5}) = (x^{1/5})^5 = x\). For the second, substitute \(f(x)\) into \(f^{-1}\): \((x^5)^{1/5} = x\). Both compositions return \(x\), verifying the inverse function.
Key Concepts
Function compositionDomain and rangeOne-to-one function
Function composition
In mathematics, the concept of function composition is integral to understanding how functions relate to one another. Function composition involves combining two functions to form a new function. Given functions \( f(x) \) and \( g(x) \), their composition is written as \((f \circ g)(x)\), which means you apply \( g \) first, then \( f \).
Function composition is useful when checking for inverse functions. If \( f \) is a function and \( f^{-1} \) is its inverse function, the composition \( f(f^{-1}(x)) \) returns \( x \), as does \( f^{-1}(f(x)) \).
Function composition is useful when checking for inverse functions. If \( f \) is a function and \( f^{-1} \) is its inverse function, the composition \( f(f^{-1}(x)) \) returns \( x \), as does \( f^{-1}(f(x)) \).
- \( f(f^{-1}(x)) = x \)
- \( f^{-1}(f(x)) = x \)
Domain and range
The domain and range are fundamental concepts when dealing with functions. The domain of a function is the complete set of possible inputs or \( x \)-values. The range is the set of possible outputs or \( y \)-values. For the function \( f(x) = x^5 \), both the domain and range are all real numbers (\( \mathbb{R} \)).
When it comes to inverse functions, determining the domain and range is crucial. Once you find an inverse function, such as \( f^{-1}(x) = x^{1/5} \), its domain becomes the range of the original function, and its range becomes the domain of the original function.
When it comes to inverse functions, determining the domain and range is crucial. Once you find an inverse function, such as \( f^{-1}(x) = x^{1/5} \), its domain becomes the range of the original function, and its range becomes the domain of the original function.
- Original Domain of \( f(x) = x^5 \) : \( \mathbb{R} \)
- Original Range of \( f(x) = x^5 \) : \( \mathbb{R} \)
- Inverse Domain of \( f^{-1}(x) = x^{1/5} \) : \( \mathbb{R} \)
- Inverse Range of \( f^{-1}(x) = x^{1/5} \) : \( \mathbb{R} \)
One-to-one function
A one-to-one function is an important type of function where each input is paired with a unique output, and no two different inputs produce the same output. This property makes it possible to find an inverse function. For a function to have an inverse, it must be one-to-one.
To determine if a function is one-to-one, you can use the Horizontal Line Test: if any horizontal line intersects the graph of the function at most once, the function is one-to-one.
For the function \( f(x) = x^5 \), it is inherently one-to-one since it is strictly increasing for all real numbers. This means each \( x \)-value produces a distinct output, allowing us to create an inverse function \( f^{-1}(x) = x^{1/5} \).
To determine if a function is one-to-one, you can use the Horizontal Line Test: if any horizontal line intersects the graph of the function at most once, the function is one-to-one.
For the function \( f(x) = x^5 \), it is inherently one-to-one since it is strictly increasing for all real numbers. This means each \( x \)-value produces a distinct output, allowing us to create an inverse function \( f^{-1}(x) = x^{1/5} \).
- One-to-one functions have inverses that are also functions.
- Ensure bijection: each input maps to a unique output.
Other exercises in this chapter
Problem 19
In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\frac{\ln t}{t} $$
View solution Problem 19
Find the derivative of \(y\) with respect to the given independent variable. \(y=7^{\sec \theta} \ln 7\)
View solution Problem 20
In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\operatorname{csch} \theta(1-\ln \operatorname{csch} \theta)
View solution Problem 20
In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=e^{\left(4 \sqrt{x}+x^{2}\right)} $$
View solution