Step 1: 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 $\mathbb{R}$. Step 2: Determine the range of the function for $x \le 0$. 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 $(-\infty, 1]$. Step 3: Determine the range of the function for $x > 0$. 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 $(1, \infty)$. Step 4: Combine the ranges from both parts. 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 $\boxed{\text{A}}$