Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the $x$-values at which they occur. \[ f(x)=4+x-x^{2} ;[0,2] \] The absolute maximum value is at $x=$ (Round to two decimal places as needed. Use a comma to separate answers as needed.)

Step 1 ：The function given is a quadratic function, and its graph is a parabola opening downwards since the coefficient of \(x^2\) is negative.

Step 2 ：The maximum or minimum of a quadratic function \(ax^2+bx+c\) occurs at \(x=-\frac{b}{2a}\).

Step 3 ：However, since we are given a specific interval [0,2], we need to evaluate the function at the endpoints of the interval and at the critical point, and then compare these values to find the absolute maximum and minimum.

Step 4 ：The critical point is where the derivative of the function is zero or undefined. In this case, the derivative of the function is \(f'(x)=1-2x\), and it is zero when \(x=\frac{1}{2}\).

Step 5 ：So, we need to evaluate the function at \(x=0\), \(x=2\), and \(x=\frac{1}{2}\).

Step 6 ：The function values at \(x=0\), \(x=\frac{1}{2}\), and \(x=2\) are 4, 4.25, and 2 respectively.

Step 7 ：The maximum value is 4.25, which occurs at \(x=\frac{1}{2}\), and the minimum value is 2, which occurs at \(x=2\).

Step 8 ：However, the question only asks for the \(x\)-value at which the maximum value occurs. So, the answer is \(x=\frac{1}{2}\).

Step 9 ：Final Answer: The absolute maximum value is at \(x=\boxed{0.50}\).