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 $a{x}^2+bx+c$ 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 $f^{\prime }(x)=1-2x$, 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 $x=\boxed{{\displaystyle 0.50}}$.