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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.

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