Use the equation and the corresponding graph for the quadratic function to find the domain and range.
\[
f(x)=-2(x-2)^{2}+1
\]

Step 1 ：The given function is a quadratic function in the form of \(f(x) = a(x-h)^2 + k\), where (h, k) is the vertex of the parabola. The coefficient 'a' determines the direction of the parabola. If 'a' is positive, the parabola opens upwards, and if 'a' is negative, the parabola opens downwards.

Step 2 ：In the given function \(f(x) = -2(x-2)^2 + 1\), the vertex of the parabola is (2, 1) and 'a' is -2, which means the parabola opens downwards.

Step 3 ：The domain of a function is the set of all possible x-values. For a quadratic function, the domain is all real numbers, because you can substitute any real number for x and get a real number for f(x). So, the domain of the function is \(D = (-∞, ∞)\) or all real numbers.

Step 4 ：The range of a function is the set of all possible y-values. Since the parabola opens downwards and the vertex is the highest point on the graph, the range of the function is all y-values less than or equal to the y-coordinate of the vertex. So, the range of the function is \(R = (-∞, 1]\) or all real numbers less than or equal to 1.

Step 5 ：So, the domain of the function \(f(x) = -2(x-2)^2 + 1\) is \(\boxed{D = (-∞, ∞)}\) and the range is \(\boxed{R = (-∞, 1]}\).