91Ó°ÊÓ

Let's first inspect the continuity of the two individual functions \(f(x) = \ln(x+3)\) and \(h(x) = \sqrt{4-x^2}\). A logarithmic function like \(f(x)\) is continuous for all values within its domain. The domain of \(f(x) = \ln(x+3)\) is \(x+3 > 0\), which means \(x > -3\). Since our interval [-2, 1] is a subset of that domain, \(f(x)\) is continuous throughout the entire interval. Now let's consider the function \(h(x) = \sqrt{4-x^2}\). The square root function is continuous for non-negative input values. In this case, the input value is \(4-x^2\). To ensure its domain falls within the interval [-2,1], we must check that \(4-x^2 \geq 0\) throughout the interval. For \(x\in[-2,1]\), this inequality holds true: \[ 4-x^2 \geq 0 \Rightarrow -x^2 \geq -4 \Rightarrow x^2 \leq 4. \] Thus, \(h(x)\) is continuous throughout the interval [-2, 1] as well. Since both \(f(x)\) and \(h(x)\) are continuous throughout the interval [-2, 1], their sum, the given function \(g(x) = f(x)+h(x) = \ln(x+3) + \sqrt{4-x^2}\), is also continuous throughout the entire interval. #Conclusion# The function \(g(x) = \ln(x+3) + \sqrt{4-x^2}\) is continuous on the closed interval [-2,1].