In Exercises 41–44:
a. Find f⁻¹(x).

44. f(x) = 2x², x ≥ 0, a = 5

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Identify the function given: \(f(x) = 2x^{2}\) with the domain restriction \(x \geq 0\). This restriction is important because it ensures the function is one-to-one and thus invertible on this domain.

To find the inverse function \(f^{-1}(x)\), start by replacing \(f(x)\) with \(y\): write \(y = 2x^{2}\).

Next, swap the roles of \(x\) and \(y\) to find the inverse: write \(x = 2y^{2}\), where now \(y\) represents the inverse function output.

Solve this equation for \(y\): divide both sides by 2 to get \(\frac{x}{2} = y^{2}\), then take the square root of both sides. Since the original domain restricts \(x \geq 0\), take the positive root: \(y = \sqrt{\frac{x}{2}}\).

Express the inverse function explicitly as \(f^{-1}(x) = \sqrt{\frac{x}{2}}\). This is the formula for the inverse function on the domain \(x \geq 0\).

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주요 개념

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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. Finding the inverse involves solving the equation y = f(x) for x in terms of y.

Domain Restriction for Invertibility

A function must be one-to-one (injective) to have an inverse. Since f(x) = 2x² is not one-to-one over all real numbers, restricting the domain to x ≥ 0 ensures the function is increasing and invertible on that interval.

Solving Quadratic Equations

To find the inverse of f(x) = 2x², you solve for x in terms of y by isolating x: y = 2x² implies x = ±√(y/2). The domain restriction x ≥ 0 selects the positive root, ensuring the inverse function is well-defined.

In Exercises 41–44:a. Find f⁻¹(x).44. f(x) = 2x², x ≥ 0, a = 5