Chapter 5: Problem 30
If \(f\) is odd and invertible, prove that \(f^{-1}\) is also odd.
Short Answer
Expert verified
If \( f \) is odd and invertible, then \( f^{-1} \) is also odd because \( f^{-1}(-y) = -f^{-1}(y) \).
Step by step solution
01
Define an Odd Function
A function \( f \) is considered odd if for all \( x \) in its domain, \( f(-x) = -f(x) \). This is the key property of odd functions.
02
Consider the Inverse of \( f \)
Since \( f \) is invertible, every element in the range of \( f \) has a unique pre-image in the domain. The inverse function \( f^{-1} \) satisfies \( f(f^{-1}(y)) = y \) for any \( y \) in the domain of \( f^{-1} \).
03
Express \( f^{-1} \) in Terms of \( f \)
From the property of the inverse function, \( f(f^{-1}(y)) = y \). Now, express \( f^{-1}(-y) \) using \( f \): since \( f \) is odd, \( f(-a) = -f(a) \), so \( f(f^{-1}(-y)) = -y \).
04
Prove \( f^{-1}(-y) = -f^{-1}(y) \)
We need to show that for \( y \) in the domain of \( f^{-1} \), \( f^{-1}(-y) = -f^{-1}(y) \). Since \( f(f^{-1}(-y)) = -y \) and by substituting \( x = f^{-1}(y) \), this implies \( f(f^{-1}(-y)) = f(-x) = -f(x) = -y \). Thus, \( f^{-1}(-y) = -x = -f^{-1}(y) \).
05
Conclude that \( f^{-1} \) is Odd
From Step 4, we have shown that \( f^{-1}(-y) = -f^{-1}(y) \) for all \( y \) in the domain of \( f^{-1} \), satisfying the definition of an odd function. Therefore, \( f^{-1} \) is indeed odd.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Invertible Function
An invertible function is a special type of function that has a unique quality: every output value maps to one and only one input value. This means that if a function is invertible, we can "reverse" it to find a corresponding input for each output. In more technical terms, a function \( f \) is invertible if there exists another function \( f^{-1} \) such that both \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \) hold true for all applicable \( x \) and \( y \). This property ensures that \( f^{-1} \) effectively "undoes" the action of \( f \), leading to a perfect mapping between inputs and outputs on both sides.
- An invertible function must be bijective, meaning it is both one-to-one and onto.
- One-to-one (injective) means that no two different inputs produce the same output.
- Onto (surjective) means that every possible output is mapped to by some input.
Inverse Function
The inverse function, denoted as \( f^{-1}(x) \), is essentially a way to "reverse" a given function \( f \). If you have an output from \( f \), the inverse function \( f^{-1} \) gives you back the original input. To find the inverse of a function, you swap the roles of the input and output. Now, for a function to have an inverse, the original function needs to be invertible, as discussed earlier.
- The relationship is described by: \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \).
- If we take any specific value of the original function's range, the inverse function maps it back to the original domain.
Properties of Odd Functions
Odd functions exhibit a unique symmetry. Mathematically, a function \( f \) is called odd if for all \( x \) in its domain, the equation \( f(-x) = -f(x) \) holds true. This symmetry can be visualized as a shape that is perfectly mirrored opposite the origin on a graph.
- This property implies that if you rotate the graph 180 degrees around the origin, it looks the same.
- A practical example would be \( f(x) = x^3 \) or \( f(x) = \sin x \).
Mathematical Proof
Mathematical proof is a logical argument that confirms the truth of a statement or theorem. These proofs help mathematicians verify properties and relationships within functions. In the context of proving whether the inverse of an odd function is also odd, we follow a structured approach involving known properties of functions.
- First, express and exploit known properties. Here, we use the definition of an odd function and properties of the inverse.
- The goal is to manipulate these properties to show that for \( f^{-1}(-y) = -f^{-1}(y) \) holds for all \( y \) in the domain of \( f^{-1} \).