Find the global maximum and minimum for the function on the closed interval. \[ f(x)=x-2 \ln (x+1) \quad 0 \leq x \leq 2 \]

Step 1 ：Find the global maximum and minimum for the function on the closed interval \(f(x)=x-2 \ln (x+1) \quad 0 \leq x \leq 2\)

Step 2 ：Take the derivative of the function: \(f'(x) = 1 - \frac{2}{x + 1}\)

Step 3 ：Find the critical points by setting the derivative to zero: \(1 - \frac{2}{x + 1} = 0\) which gives \(x = 1\)

Step 4 ：Evaluate the function at the critical points and the endpoints of the interval: \(f(0) = 0\), \(f(1) = 1 - 2 \ln(2)\), and \(f(2) = 2 - 2 \ln(3)\)

Step 5 ：Compare the values to determine the global maximum and minimum: \(\boxed{\text{Global Maximum: }(0, 0)}\) and \(\boxed{\text{Global Minimum: }(1, 1 - 2 \ln(2))}\)