Problem 58
Question
If a composition \(f^{\circ} g\) is one-to-one, must \(g\) be one-to-one? Give reasons for your answer.
Step-by-Step Solution
Verified Answer
Yes, if \(f^{\circ} g\) is one-to-one, then \(g\) must be one-to-one.
1Step 1: Understand the Problem
We are asked to determine whether the function \(g\) must be one-to-one if the composition \(f^{\circ} g\) (i.e., \(f(g(x))\)) is one-to-one. This requires understanding what it means for a function to be one-to-one: a function \(h\) is one-to-one if, for every pair of different inputs, the outputs are different, i.e., \(h(a) eq h(b)\) whenever \(a eq b\).
2Step 2: Recall Definitions
Recall that the composition of two functions \(f^{\circ} g\) is defined as \(f(g(x))\). A function is considered one-to-one (injective) if no two different elements in the domain map to the same element in the codomain.
3Step 3: Analyze the Composition
Consider function \(f\) and \(g\), where the composition \(f(g(x))\) is one-to-one. If \(g(x_1) = g(x_2)\) for any \(x_1 eq x_2\), then \(f(g(x_1)) = f(g(x_2))\). This contradicts the assumption that \(f(g(x))\) is one-to-one. Thus, \(g\) must map different inputs to different outputs, suggesting \(g\) must be one-to-one.
4Step 4: Conclude
Since we derived that \(g(x_1) = g(x_2)\) leads to a contradiction when \(f(g(x))\) is known to be one-to-one, we conclude that \(g\) must truly map distinct values to distinct outputs. Therefore, if \(f^{\circ} g\) is one-to-one, then \(g\) must be one-to-one.
Key Concepts
Composition of FunctionsInjective FunctionFunction MappingFunction Domain
Composition of Functions
When we talk about the composition of functions, we are combining two functions to create a new function. If we have two functions, say \( f \) and \( g \), the composition is written as \( f \circ g \) or \( f(g(x)) \). This means that you first apply the function \( g \) to \( x \), and then apply the function \( f \) to the result.
The importance of the composition lies in how it combines processes: like first preparing ingredients with \( g \) before cooking them with \( f \). It is crucial to follow the order precisely for the desired outcome, just like following a recipe.
In mathematics, the order of operations can change the result, so understanding the sequence in function compositions is essential.
The importance of the composition lies in how it combines processes: like first preparing ingredients with \( g \) before cooking them with \( f \). It is crucial to follow the order precisely for the desired outcome, just like following a recipe.
In mathematics, the order of operations can change the result, so understanding the sequence in function compositions is essential.
- In \( f(g(x)) \), always follow: apply \( g \) first, then \( f \).
- The composition reflects a chain, where the output of \( g \) becomes the input to \( f \).
Injective Function
An injective function, also known as a one-to-one function, is a special type of function where each element in the function's domain is mapped to a unique element in the codomain.
In simple terms, different inputs will never map to the same output. This is a key feature in ensuring that a function's output is reliable and distinct.
You can visualize this by imagining a room full of unique keys (inputs) where each key fits into only one specific lock (output).
How to identify an injective function? Here are some pointers:
In simple terms, different inputs will never map to the same output. This is a key feature in ensuring that a function's output is reliable and distinct.
You can visualize this by imagining a room full of unique keys (inputs) where each key fits into only one specific lock (output).
How to identify an injective function? Here are some pointers:
- If \( h(a) eq h(b) \) whenever \( a eq b \), then \( h \) is injective.
- No two different elements from the domain can have the same image in the codomain.
Function Mapping
Function mapping describes the relationship between a function’s input and its corresponding output. It serves as a blueprint for understanding how each element from one set relates to an element in another.
Think of a map guiding you from your house to school—it tells you exactly where you start and where you'll end up.
Here are some highlights of function mapping:
Think of a map guiding you from your house to school—it tells you exactly where you start and where you'll end up.
Here are some highlights of function mapping:
- It demonstrates how elements of the domain correspond to elements in the codomain.
- For a function \( f \colon A \rightarrow B \), every element in set \( A \) (domain) is paired with an element in set \( B \) (codomain).
Function Domain
The domain of a function refers to all possible input values (or x-values) for which the function is defined. In simpler words, it's the set of all inputs that a function can handle without causing any errors or undefined behavior.
Why is the domain important?
For example, if a function involves a square root, the domain would include only non-negative numbers as inputs to keep the outputs real. Similarly, if the denominator in a fraction is a variable, the domain cannot include values that make the denominator zero.
Knowing the domain avoids mistakes and ensures that every input has a meaningful and valid output.
Why is the domain important?
- The domain sets the boundaries for which a function can operate correctly and logically.
- Understanding the domain prevents calculations from involving numbers or scenarios that a function can't handle.
For example, if a function involves a square root, the domain would include only non-negative numbers as inputs to keep the outputs real. Similarly, if the denominator in a fraction is a variable, the domain cannot include values that make the denominator zero.
Knowing the domain avoids mistakes and ensures that every input has a meaningful and valid output.
Other exercises in this chapter
Problem 58
Solve the initial value problems. \begin{equation}\frac{d^{2} y}{d t^{2}}=1-e^{2 t}, \quad y(1)=-1 \quad \text { and } \quad y^{\prime}(1)=0 \end{equation}
View solution Problem 58
In Exercises \(57-70\) , use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt{\left(x^{2}+1
View solution Problem 59
Evaluate the integrals in Exercises \(47-70\) $$ \int \frac{3 d r}{\sqrt{1-4(r-1)^{2}}} $$
View solution Problem 59
Find the derivative of \(y\) with respect to the given independent variable. \begin{equation}y=2^{x}\end{equation}
View solution