Problem 12
Question
Is the inverse of a one-to-one function always a function?
Step-by-Step Solution
Verified Answer
Yes, the inverse of a one-to-one function is always a function.
1Step 1: Understanding a One-to-One Function
A one-to-one function, also known as an injective function, is defined as a function where each element in the domain is mapped to a unique element in the range. This means that if \( f(a) = f(b) \) then \( a = b \). In simpler terms, no two different input values will yield the same output.
2Step 2: Defining the Inverse of a Function
The inverse of a function \( f(x) \), denoted as \( f^{-1}(x) \), is a function that reverses the roles of the domain and range of the original function. This means if \( f(x) = y \), then \( f^{-1}(y) = x \). For an inverse to exist, each output \( y \) must correspond to exactly one input \( x \).
3Step 3: Determining When an Inverse is a Function
For the inverse of a function to also be a function, it must pass the horizontal line test. This is equivalent to saying the original function must be one-to-one (injective), meaning each value in the function’s range must map back to only one value in its domain. This ensures that \( f^{-1}(y) \) is well-defined and meets the definition of a function.
4Step 4: Conclusion
Since a one-to-one function by definition ensures unique mapping in both directions (i.e., one input for one output and vice versa), the inverse of a one-to-one function is guaranteed to be a function. Thus, the statement is true: the inverse of a one-to-one function is always a function.
Key Concepts
Inverse FunctionInjective FunctionHorizontal Line Test
Inverse Function
An inverse function is like finding the reverse path on a journey. To compute an inverse for a function, you're essentially swapping the roles of inputs and outputs. If you have a function \( f(x) \) that turns \( x \) into \( y \), the inverse function \( f^{-1}(x) \) will do the opposite by taking \( y \) and returning you to \( x \). This requires a particular quality: every output must lead back to one single input.
To illustrate, consider the function \( f(x) = 2x+3 \). Solving for \( x \) in terms of \( y \) gives us the inverse function. So if \( f(x) = y \), reconfiguring to solve for \( x \) gives \( f^{-1}(x) = (x-3)/2 \). Hence, the inverse swaps the role of inputs and outputs and is itself a function when the original is one-to-one.
This careful pairing ensures that when we apply the inverse, we're retracing our steps correctly without ambiguity.
To illustrate, consider the function \( f(x) = 2x+3 \). Solving for \( x \) in terms of \( y \) gives us the inverse function. So if \( f(x) = y \), reconfiguring to solve for \( x \) gives \( f^{-1}(x) = (x-3)/2 \). Hence, the inverse swaps the role of inputs and outputs and is itself a function when the original is one-to-one.
This careful pairing ensures that when we apply the inverse, we're retracing our steps correctly without ambiguity.
Injective Function
Also known as one-to-one, an injective function means no two distinct inputs send you crashing into the same output. This quality is crucial for ensuring an inverse can exist as a function.
Why does being injective matter? Because when a function is injective, it guarantees unique outputs for every unique input. Imagine a crowd of people where each person has a distinct name. In contrast, in a non-injective situation, like a generic name label, two or more people might end up claiming the same name.
- **Injectivity** asserts that: - If \( f(a) = f(b) \), then \( a = b \) must also hold.- No two different inputs \( a \) and \( b \) can lead to the same result.
By validating this exclusivity, injecting ensures every output is traceable back to a single, unique input, paving the way for a clear, inverse relation.
Why does being injective matter? Because when a function is injective, it guarantees unique outputs for every unique input. Imagine a crowd of people where each person has a distinct name. In contrast, in a non-injective situation, like a generic name label, two or more people might end up claiming the same name.
- **Injectivity** asserts that: - If \( f(a) = f(b) \), then \( a = b \) must also hold.- No two different inputs \( a \) and \( b \) can lead to the same result.
By validating this exclusivity, injecting ensures every output is traceable back to a single, unique input, paving the way for a clear, inverse relation.
Horizontal Line Test
The horizontal line test is a simple visual check to decide if a function is one-to-one, hence invertible. Imagine drawing horizontal lines across a function's graph. If any line touches the graph more than once, the function fails the test.
Visualize the journey of a function's graph; a horizontal line must intersect each path at most once. If it pierces more than once, it shows overlapping outputs for distinct inputs, meaning no inverse function can truly exist.
This test is crucial:- It confirms if a function is injective.- For example, the quadratic function \( f(x) = x^2 \) fails as many horizontal lines will cross it twice. Thus, it isn't invertible over all real numbers.
The horizontal line test is a quick, effective method to ascertain invertibility, ensuring clarity and isolating functions that are sharp and distinct enough to have tidy inverses.
Visualize the journey of a function's graph; a horizontal line must intersect each path at most once. If it pierces more than once, it shows overlapping outputs for distinct inputs, meaning no inverse function can truly exist.
This test is crucial:- It confirms if a function is injective.- For example, the quadratic function \( f(x) = x^2 \) fails as many horizontal lines will cross it twice. Thus, it isn't invertible over all real numbers.
The horizontal line test is a quick, effective method to ascertain invertibility, ensuring clarity and isolating functions that are sharp and distinct enough to have tidy inverses.
Other exercises in this chapter
Problem 12
Fill in the blanks. The functions \(f(x)=\log _{10} x\) and \(f(x)=10^{x}\) are ____ functions.
View solution Problem 12
Match expression with an equivalent expression from the list on the right. \(\log _{3} \frac{10}{11}\) a. \(\frac{\log 11}{\log 3}\) b. \(11 \log _{3} 10\) c. \
View solution Problem 13
a. How do we solve \(x \ln 3=\ln 5\) for \(x ?\) b. What is the exact solution? c. What is an approximate solution to four decimal places?
View solution Problem 13
Let \(f(x)=2 x+1\) and \(g(x)=x-3 .\) Find each function and give its domain. $$ f+g $$
View solution