Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.


f(x) = 1/x², x > 0

Verified step by step guidance

1. Start by writing the function given: \(f(x) = \frac{1}{x^{2}}\) with the domain \(x > 0\).
2. To find the inverse function \(f^{-1}(x)\), set \(y = \frac{1}{x^{2}}\) and then interchange \(x\) and \(y\) to get \(x = \frac{1}{y^{2}}\).
3. Solve the equation \(x = \frac{1}{y^{2}}\) for \(y\): multiply both sides by \(y^{2}\) to get \(x y^{2} = 1\), then divide by \(x\) to get \(y^{2} = \frac{1}{x}\).
4. Take the square root of both sides to solve for \(y\): \(y = \pm \sqrt{\frac{1}{x}} = \pm \frac{1}{\sqrt{x}}\). Since the original domain of \(f\) is \(x > 0\) and \(f(x) > 0\), choose the positive root to maintain the inverse function's domain and range consistency, so \(f^{-1}(x) = \frac{1}{\sqrt{x}}\).
5. Identify the domain and range of \(f^{-1}\): the domain of \(f^{-1}\) is the range of \(f\), which is \(y > 0\), so \(x > 0\) for \(f^{-1}\). The range of \(f^{-1}\) is the domain of \(f\), which is \(y > 0\). Finally, verify the inverse by checking that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).

Key Concepts

Inverse Functions

An inverse function reverses the effect of the original function, swapping inputs and outputs. For a function f(x), its inverse f⁻¹(x) satisfies f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. Finding the inverse involves solving y = f(x) for x in terms of y.

Domain and Range of Functions and Their Inverses

The domain of a function is the set of all possible input values, while the range is the set of possible outputs. For an inverse function, the domain and range swap roles compared to the original function. Identifying these sets ensures the inverse is well-defined.

Restrictions on the Domain for Invertibility

A function must be one-to-one (injective) to have an inverse. Restricting the domain, such as x > 0 for f(x) = 1/x², ensures the function is one-to-one and invertible. Without this, the inverse would not be a proper function.