Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the minimum value of \(a\), we first need to understand the given functions \(f(x)\) and \(g(x)\) and how they relate to each other through the given inequality \(f(x_1) \geq g(x_2)\) for \(x_1 \in [0,1]\) and \(x_2 \in [1,2]\). The function \(f(x) = x - \dfrac{1}{x+1}\) and \(g(x) = x^2 - 2ax + 4\), where \(a\) is a parameter that we need to minimize.</p> <p><strong>Step 2:</strong> We start by analyzing \(f(x)\) over the interval \([0,1]\). Since \(f(x) = x - \dfrac{1}{x+1}\), let's find its minimum and maximum values in this interval to understand the range of \(f(x_1)\). To do this, we can differentiate \(f(x)\) with respect to \(x\), yielding \(f'(x) = 1 + \dfrac{1}{(x+1)^2}\), which is always positive, indicating \(f(x)\) is increasing over \([0,1]\). Thus, the minimum value of \(f(x_1)\) occurs at \(x_1 = 0\) and the maximum at \(x_1 = 1\). Calculating these: \(f(0) = 0 - \dfrac{1}{0+1} = -1\) and \(f(1) = 1 - \dfrac{1}{1+1} = \dfrac{1}{2}\).</p> <p><strong>Step 3:</strong> Next, we consider \(g(x_2) = x_2^2 - 2ax_2 + 4\) for \(x_2 \in [1,2]\). To ensure \(f(x_1) \geq g(x_2)\) holds for all \(x_1\) in \([0,1]\), we need the maximum value of \(g(x_2)\) over \([1,2]\) to be less than or equal to the minimum value of \(f(x_1)\), which is \(-1\), or we find a condition on \(a\) that satisfies the given inequality for the range of \(f(x_1)\) and \(g(x_2)\). Since \(g(x)\) is a quadratic function, its maximum or minimum occurs at its vertex, \(x = a\). However, since we're constrained to \(x_2 \in [1,2]\), we must evaluate \(g(x_2)\) at the endpoints of this interval and potentially at its vertex if \(a \in [1,2]\).</p> <p><strong>Step 4:</strong> Evaluating \(g(x_2)\) at \(x_2 = 1\) and \(x_2 = 2\), we get \(g(1) = 1 - 2a + 4 = 5 - 2a\) and \(g(2) = 4 - 4a + 4 = 8 - 4a\). To satisfy the condition that \(f(x_1) \geq g(x_2)\) for all \(x_1 \in [0,1]\) and some \(x_2 \in [1,2]\), considering the minimum of \(f(x_1)\) is \(-1\), we require that the maximum value of \(g(x_2)\) in the interval \([1,2]\) is less than or equal to \(-1\), or we find the condition on \(a\) that makes this true.</p> <p><strong>Step 5:</strong> Considering \(g(1)\) and \(g(2)\), and knowing that \(g(x)\) could have its vertex in \([1,2]\) depending on \(a\), we aim to ensure \(g(x_2) \leq -1\) for the maximum value of \(g(x_2)\) in \([1,2]\). This implies \(5 - 2a \leq -1\) and \(8 - 4a \leq -1\), simplifying to \(2a \geq 6\) and \(4a \geq 9\), which gives \(a \geq 3\) and \(a \geq \dfrac{9}{4}\), respectively. The stricter condition to ensure \(f(x_1) \geq g(x_2)\) for all \(x_1\)