Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the value of \( \dfrac{g(0)}{g(2)} \), we first need to understand and possibly simplify the given functions \( f(x) \) and \( g(x) \). The function \( f(x) \) is defined as \( f(x) = \int_0^1 e^{x+t} f(t)\, dt \), and \( g(x) \) is defined as \( g(x) = x + \int_0^1 e^{x+t} g(t)\, dt \).</p> <p><strong>Step 2:</strong> Let's start by analyzing \( f(x) \). By substituting \( f(x) \) into its own definition, we see that \( f(x) \) is a solution to an integral equation. However, without a specific form, directly solving for \( f(x) \) or \( g(x) \) might be complex. Instead, we notice that the integral part of both \( f(x) \) and \( g(x) \) involves \( e^{x+t} \) multiplied by the function itself, integrated over \( t \) from 0 to 1. This suggests a relationship between \( f(x) \) and \( g(x) \) through their integral forms.</p> <p><strong>Step 3:</strong> To simplify the problem, let's evaluate \( g(x) \) at specific points, namely \( g(0) \) and \( g(2) \), as these are the values we are interested in. For \( g(0) \), we substitute \( x = 0 \) into the equation for \( g(x) \): \( g(0) = 0 + \int_0^1 e^{0+t} g(t)\, dt = \int_0^1 e^t g(t)\, dt \). For \( g(2) \), substituting \( x = 2 \) gives \( g(2) = 2 + \int_0^1 e^{2+t} g(t)\, dt \).</p> <p><strong>Step 4:</strong> Now, let's find a relationship that can help us solve for \( \dfrac{g(0)}{g(2)} \) without explicitly solving for \( g(t) \). Notice that \( g(2) \) can be rewritten as \( g(2) = 2 + e^2 \int_0^1 e^t g(t)\, dt \). Recognizing that \( \int_0^1 e^t g(t)\, dt = g(0) \), we have \( g(2) = 2 + e^2 g(0) \).</p> <p><strong>Step 5:</strong> To find \( \dfrac{g(0)}{g(2)} \), we rearrange the equation from Step 4: \( g(2) = 2 + e^2 g(0) \) implies \( g(0) = \dfrac{g(2) - 2}{e^2} \). Thus, \( \dfrac{g(0)}{g(2)} = \dfrac{g(2) - 2}{e^2 g(2)} \). However, we need a direct relationship or value to proceed, which we can obtain by considering the nature of \( g(x) \) and its integral form.</p> <p><strong>Step 6:</strong> Since \( g(x) = x + \int_0^1 e^{x+t} g(t)\, dt \), let's evaluate this at \( x = 0 \) and \( x = 2 \) to establish a direct relationship for \( \dfrac{g(0)}{g(2)} \). For \( g(0) \), we have \( g(0) = 0 + \int_0^1 e^t g(t)\, dt \), and for \( g(2) \), \( g(2) = 2 + \int_0^1 e^{2+t} g(t)\, dt = 2 + e^2 \int_0^1 e^t g(t)\, dt \). This implies \( g(2) = 2 + e^2 g(0) \), because \( g(0) = \int_0^1 e^t g(t)\, dt \).</p> <p><strong>Step 7:</strong> From the equation \( g(2) = 2 + e^2 g(0) \), we solve for \( g(0) \) in terms of \( g(2)