Determine the domain of the function.

The function $f(x)$ is defined in two parts:

• $f(x) = 1-x^2$ for $x \le 0$. This covers all non-positive real numbers.
• $f(x) = 2x+1$ for $x > 0$. This covers all positive real numbers. Together, these two conditions cover all real numbers. Therefore, the domain of $f(x)$ is $(-\infty, \infty)$, which is denoted by $\mathbb{R}$.

For $x \le 0$, $f(x) = 1-x^2$.
This is a parabola opening downwards with its vertex at $(0, 1)$.
When $x=0$, $f(0) = 1 - (0)^2 = 1$.
As $x$ decreases from 0 (e.g., $x=-1, -2, \dots$), $x^2$ increases, so $1-x^2$ decreases.
For example, $f(-1) = 1 - (-1)^2 = 0$, $f(-2) = 1 - (-2)^2 = -3$.
Thus, for $x \le 0$, the range of $f(x)$ is $(-\infty, 1]$.

For $x > 0$, $f(x) = 2x+1$.
This is a linear function with a positive slope.
As $x$ approaches 0 from the right, $f(x)$ approaches $2(0)+1 = 1$. Since $x > 0$, the value 1 is not included in this part's range.
As $x$ increases from 0, $f(x)$ increases.
For example, $f(1) = 2(1)+1 = 3$, $f(2) = 2(2)+1 = 5$.
Thus, for $x > 0$, the range of $f(x)$ is $(1, \infty)$.

The range of $f(x)$ is the union of the ranges from Step 2 and Step 3:
Range $= (-\infty, 1] \cup (1, \infty)$.
This union covers all real numbers.
Therefore, the range of $f(x)$ is $(-\infty, \infty)$, which is denoted by $\mathbb{R}$.

The domain is $\mathbb{R}$ and the range is $\mathbb{R}$.
Comparing this with the given options, option A states "R and R".

The final answer is $\text{A}$