Problem 47
Question
Suppose that the range of \(g\) lies in the domain of \(f\) so that the composite \(f \circ g\) is defined. If \(f\) and \(g\) are one-to-one, can anything be said about \(f \circ g ?\) Give reasons for your answer.
Step-by-Step Solution
Verified Answer
The composite function \(f \circ g\) is also one-to-one.
1Step 1: Understanding One-to-One Functions
A function is one-to-one if each element of the range is mapped by exactly one element of the domain. This means there are no repeated values in the output, therefore ensuring that each input gives a unique output.
2Step 2: Composite Function Definition
The composite function \((f \circ g)(x)\) is defined as \(f(g(x))\). For \(f \circ g\) to be defined, the range of \(g\) must be a subset of the domain of \(f\).
3Step 3: Check if Composite Function is One-to-One
For the composite function \(f \circ g\) to be one-to-one, it should map each element of its domain to a unique element in its range. Since \(g\) is one-to-one, \(g(x) eq g(y)\) implies \(x eq y\). Likewise, since \(f\) is one-to-one, \(f(a) eq f(b)\) implies \(a eq b\). Therefore, the composite \(f(g(x)) eq f(g(y))\) holds true whenever \(x eq y\), making \(f \circ g\) one-to-one.
Key Concepts
One-to-One FunctionsComposite FunctionsFunction Range and Domain
One-to-One Functions
A one-to-one (injective) function is a special type of function where each element in the domain maps to a unique element in the range. This means that no two different inputs produce the same output. To recognize one-to-one functions, recall these key points:
- If a function is one-to-one, then whenever two elements in the domain have the same image in the range, they must be the same element. Mathematically, this means if \( f(a) = f(b) \), then \( a = b \).
- Graphically, a function is one-to-one if any horizontal line drawn across its graph does not intersect the graph more than once. This is known as passing the "horizontal line test."
Composite Functions
Composite functions involve combining two functions such that the output of one function becomes the input of another. The notation \((f \circ g)(x)\) represents the composite function defined by \( f(g(x)) \). Here's how composite functions work:
- The function \( g \) is applied first to \( x \), and then the result \( g(x) \) is used as the input for the function \( f \).
- To ensure that \( f \circ g\) is defined, the range of \( g \) (outputs of \( g \)) must lie entirely within the domain of \( f \) (inputs for \( f \)).
- The process can be visualized as a two-step journey: from \( x \) to \( g(x) \), and from there to \( f(g(x)) \), creating a seamless flow from start to finish.
Function Range and Domain
Understanding the concepts of domain and range is crucial when working with functions. The domain of a function is the set of all possible inputs, while the range is the set of all possible outputs produced by the function.
- The domain of a composite function \( f \circ g \) is influenced by the domains of the individual functions \( f \) and \( g \). For the composite function to be valid:
- You must be able to input any value from \( g \)'s range into \( f \). Hence, \( g(x) \) must fall within the domain of \( f \).
- The range of a composite function depends partly on the range of \( g \) and the way \( f \) transforms the inputs it receives from \( g \). Each transformation leads to the final outcomes or values in the composite range \( f(g(x)) \).
Other exercises in this chapter
Problem 47
Evaluate the integrals in Exercises \(37-54\). $$ \int \frac{3 \sec ^{2} t}{6+3 \tan t} d t $$
View solution Problem 47
Evaluate the integrals. \(\int 5^{x} d x\)
View solution Problem 48
Evaluate the integrals in Exercises \(41-50\) . $$ \int \operatorname{csch}^{2}(5-x) d x $$
View solution Problem 48
Find the limits in Exercises \(41-48 .\) (If in doubt, look at the function's graph.) $$ \lim _{x \rightarrow-\infty} \csc ^{-1} x $$
View solution