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 $R$, determine if $R$ defines a function from $\mathbb{R}$ to $\mathbb{R}$, then define a function $f$ by restricting $R$, find its inverse $f^{-1}$, and analyze if $f^{-1}$ is a function. 2. **Find the domain of $R$:** The relation is defined by $$y^2 = 4x - x^2.$$ For $y^2$ to be real and non-negative, the right side must satisfy $$4x - x^2 \geq 0.$$ Factor: $$4x - x^2 = x(4 - x) \geq 0.$$ This inequality holds when $x \in [0,4]$ because: - For $x < 0$, $x$ is negative and $4-x > 4$, so product negative. - For $x > 4$, $4-x$ is negative, so product negative. Thus, the domain is $$\boxed{[0,4]}.$$ 3. **Find the range of $R$:** From $$y^2 = 4x - x^2,$$ the maximum value of $y^2$ occurs at the vertex of the quadratic in $x$: Rewrite as $$-x^2 + 4x = -(x^2 - 4x) = -(x^2 - 4x + 4) + 4 = -(x-2)^2 + 4.$$ The maximum is 4 at $x=2$, so $$y^2 \leq 4 \implies |y| \leq 2.$$ Hence, the range is $$\boxed{[-2,2]}.$$ 4. **Does $R$ define a function from $\mathbb{R}$ to $\mathbb{R}$?** For each $x$ in the domain, there are two possible $y$ values: $$y = \pm \sqrt{4x - x^2}.$$ Since there are two $y$ values for some $x$ (except at $x=0$ and $x=4$), $R$ is not a function. 5. **Define a function $f$ by restricting $R$:** To make $R$ a function, restrict to the upper half (non-negative $y$): $$f(x) = \sqrt{4x - x^2}, \quad x \in [0,4].$$ 6. **Find the inverse $f^{-1}$:** Start with $$y = \sqrt{4x - x^2}.$$ Square both sides: $$y^2 = 4x - x^2.$$ Rearrange: $$x^2 - 4x + y^2 = 0.$$ Treating $x$ as variable: $$x^2 - 4x + y^2 = 0.$$ Use quadratic formula: $$x = \frac{4 \pm \sqrt{16 - 4y^2}}{2} = 2 \pm \sqrt{4 - y^2}.$$ Since $f$ has domain $[0,4]$ and range $[0,2]$, and $f$ is increasing on $[0,2]$ and decreasing on $[2,4]$, the inverse function must be defined on $[0,2]$. To get a function inverse, restrict $f$ to $[0,2]$ where it is increasing: - For $x \in [0,2]$, $f(x) = \sqrt{4x - x^2}$ is increasing. Then inverse is $$f^{-1}(y) = 2 - \sqrt{4 - y^2}, \quad y \in [0,2].$$ 7. **Is $f^{-1}$ a function?** Yes, with this restriction, $f^{-1}$ is a function from $[0,2]$ to $[0,2]$. 8. **Summary:** - Domain of $R$: $[0,4]$ - Range of $R$: $[-2,2]$ - $R$ is not a function. - Restrict $R$ to $f(x) = \sqrt{4x - x^2}$ on $[0,4]$ to get a function. - Restrict domain further to $[0,2]$ to get inverse function $$f^{-1}(y) = 2 - \sqrt{4 - y^2}.$$