1. **Problem statement:** Given the relation $R = \{(x,y) \in \mathbb{R}^2 \mid y^2 = 4x - x^2\}$, find the domain and range of $R$ and determine if $R$ defines a function from $\mathbb{R}$ to $\mathbb{R}$. 2. **Formula and rules:** The relation is defined by the equation $y^2 = 4x - x^2$. To find the domain, we find all $x$ such that the right side is nonnegative (since $y^2 \geq 0$). To find the range, we find all possible $y$ values. A relation defines a function if for every $x$ there is exactly one $y$. 3. **Find domain:** $$y^2 = 4x - x^2 \geq 0$$ Rewrite as $$4x - x^2 = -x^2 + 4x = -(x^2 - 4x) = -(x(x-4)) \geq 0$$ The inequality $-(x(x-4)) \geq 0$ means $$x(x-4) \leq 0$$ This holds when $x$ is between 0 and 4 inclusive. So, $$\text{Domain} = [0,4]$$ 4. **Find range:** From $y^2 = 4x - x^2$, the maximum of $4x - x^2$ on $[0,4]$ occurs at $x=2$ (vertex of parabola). Calculate $$4(2) - (2)^2 = 8 - 4 = 4$$ So, $$y^2 \leq 4 \implies |y| \leq 2$$ Hence, $$\text{Range} = [-2,2]$$ 5. **Is $R$ a function?** For each $x$ in $[0,4]$, $y$ satisfies $$y = \pm \sqrt{4x - x^2}$$ There are two $y$ values for most $x$ except at $x=0$ and $x=4$. Therefore, $R$ does not define a function from $\mathbb{R}$ to $\mathbb{R}$ because it fails the vertical line test. **Final answer:** Domain: $[0,4]$ Range: $[-2,2]$ $R$ is not a function from $\mathbb{R}$ to $\mathbb{R}$ because for some $x$ there are two $y$ values.