Problem 132 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

If \(f\) is a one-to-one odd function, its inverse \(f^{-1}\) is also an odd function.

Step 1: Declare that \(f\) is one-to-one and odd

Assume that \(f\) is a one-to-one odd function. This means that \(f(-x) = -f(x)\) for any \(x\) within the domain of \(f\), and if \(f(a) = f(b)\), then \(a = b\).

Step 2: Analyze the odd function property of \(f^{-1}\)

The goal here is to show that \(f^{-1}\) holds the property of an odd function which is \(f^{-1}(-y) = -f^{-1}(y)\). It needs to be demonstrated that this statement is true for all \(y\) within the range of \(f\).

Step 3: Use the definition of the inverse function

Substitute \(f(-x)\) for \(y\) in the formula \(f^{-1}(-y) = -f^{-1}(y)\). This gives \(f^{-1}(-f(-x)) = -f^{-1}(f(-x))\). Since \(f\) is one-to-one, where \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), it can be seen here that \(f(-x)\) is replaced with y, hence you have \(f^{-1}(-y) = -x\).

Step 4: Apply the property of the odd function

Use the definition of the odd function \(f(-x) = -f(x)\). Substituting \(f(x)\) for y in \(f^{-1}(-y) = -x\), we get \(f^{-1}(-f(x)) = -x\), which applying the property of the inverse function \(f^{-1}(f(x)) = x\) gives \(-x = -x\), satisfying the condition for the inverse to be an odd function.

Step 5: State that \(f^{-1}\) is odd

From steps 3 and 4, it has been shown that \(f^{-1}\) holds the property that the negative of its input equals the output of the negative input. Therefore, \(f^{-1}\) is an odd function.