2) Consider the function \( f(x)=2 x^{2} \) and \( p(x)=\left(\frac{1}{2}\right)^{x} \). a) Restrict the domain of \( f \) so that the inverse of \( f \) will also be a function. b) Write down the equation of the inverse of \( p \) in the form \( p^{-1}(x)=\ldots \) c) If the inverse of another function is \( g^{-1}(x)=2 x-4 \). Determine the equation of \( g(x) \).

We are given the functions f(x) = 2x² and p(x) = (1/2)ˣ. We wish to: a) Restrict the domain of f so its inverse is a function. b) Find the inverse of p in the form p⁻¹(x) = … c) Given g⁻¹(x) = 2x – 4, find g(x). Let’s work through each part step by step. ──────────────────────── Part (a) For a function to have an inverse that is also a function, the function must be one-to-one (injective). The function f(x) = 2x² is not one-to-one over the real numbers because, for example, f(2) = 8 and f(–2) = 8. A common remedy is to restrict the domain so that the function is either only increasing or only decreasing. A common restriction is to choose x ≥ 0, which gives a monotonically increasing function. Thus, by restricting the domain to [0, ∞), f becomes one-to-one and its inverse is defined. ──────────────────────── Part (b) We wish to find the inverse of p(x) = (1/2)ˣ. The steps are: 1. Begin with y = (1/2)ˣ. 2. In order to solve for x, take the logarithm base (1/2) on both sides:   x = log₍₁/₂₎(y). Since we want the inverse function written as p⁻¹(x), we interchange x and y:   p⁻¹(x) = log₍₁/₂₎(x). Thus, the inverse is:   p⁻¹(x) = log₍₁/₂₎(x). ──────────────────────── Part (c) We are given that g⁻¹(x) = 2x – 4 and need to find the corresponding function g(x). Recall that g and g⁻¹ are inverses, so:   g(g⁻¹(x)) = x. However, a more straightforward way is to “invert” the function g⁻¹(x). Write   y = 2x – 4 and solve for x in terms of y. Step 1. Start with:   y = 2x – 4 Step 2. Add 4 to both sides:   y + 4 = 2x Step 3. Divide both sides by 2:   x = (y + 4)/2 Now, reinterpret this equation. Since we began with y = g⁻¹(x) and ended with x in terms of y, the inverse of g⁻¹ is the function g. Renaming y back as the variable x, we get:   g(x) = (x + 4)/2. ──────────────────────── Final Answers: a) Restrict the domain of f to x ≥ 0. b) p⁻¹(x) = log₍₁/₂₎(x). c) g(x) = (x + 4)/2.