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Chapter 1: Problem 100

Prove that if \(f\) is a one-to-one odd function, then \(f^{-1}\) is an odd function.

Short Answer

Expert verified
In this proof, the core steps involve demonstrating the properties of an odd function and using these properties for the inverse function. After showing that \(f^{-1}(-y)\) achieves the desired condition for odd functions, we conclude that if \(f\) is a one-to-one odd function, then the inverse function \(f^{-1}\) is also odd.

Step by step solution

01

Understand the given and use the property of odd function

The main property of an odd function is that \(f(-x) = -f(x)\) for all x in the domain of \(f\). Begin by showing the similar property for \(f^{-1}\) which means we must show that \(f^{-1}(-y) = -f^{-1}(y)\). The value \(f^{-1}(-y)\) corresponds to some \(x\) such that \(f(x) = -y\).
02

Apply the property of odd function for x

Since \(f\) is an odd function, \(f(-x) = -f(x)\). Consequently, we can substitute \(f(-x)\) with \(-f(x)\).
03

Use the definition of the inverse function

Realize that \(-f(x)\) is equivalent to \(-y\). Hence, for that \(x\), the equality \(-f(x) = -y = f(-x)\) is satisfied. Therefore, \(-x\) is the value that is mapped to \(-y\) by \(f\), that is \(f^{-1}(-y) = -x\). Now take negative of both sides to get \(-f^{-1}(-y) = x\).
04

Wrapping up the proof

Since \(x = f^{-1}(y)\), \(-f^{-1}(-y) = f^{-1}(y)\). And this is exactly the definition of the odd function, namely \(f^{-1}(-y) = -f^{-1}(y)\) for all \(y\). Thus, \(f^{-1}\) is an odd function.

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