1. **Problem statement:** Determine if each relation defines a function from $\mathbb{R}$ to $\mathbb{R}$. If not, explain why and suggest domain restrictions. 2. **Recall:** A function from $\mathbb{R}$ to $\mathbb{R}$ assigns exactly one output for each input in its domain. 3. **Relation i:** $f(x) = \sqrt{x^2 - 9}$ - The expression under the square root must be non-negative: $$x^2 - 9 \geq 0$$ - Solve inequality: $$x^2 - 9 \geq 0 \implies (x-3)(x+3) \geq 0$$ - This holds when $x \leq -3$ or $x \geq 3$. - So domain restriction: $$(-\infty, -3] \cup [3, \infty)$$ - For these $x$, $f(x)$ is real and unique, so $f$ is a function on this restricted domain. 4. **Relation ii:** $f(x) = \frac{1}{x^2 - 4x + 3}$ - Denominator cannot be zero: $$x^2 - 4x + 3 = 0$$ - Factor: $$ (x-1)(x-3) = 0 $$ - So $x \neq 1$ and $x \neq 3$. - Domain restriction: $$\mathbb{R} \setminus \{1,3\}$$ - On this domain, $f$ is defined and unique, so it is a function. 5. **Relation iii:** $f(x) = \sqrt{9 - |x|}$ - Expression under root must be non-negative: $$9 - |x| \geq 0 \implies |x| \leq 9$$ - So domain restriction: $$[-9,9]$$ - For these $x$, $f(x)$ is real and unique, so $f$ is a function. **Final answers:** - (i) Function if domain restricted to $(-\infty, -3] \cup [3, \infty)$. - (ii) Function if domain restricted to $\mathbb{R} \setminus \{1,3\}$. - (iii) Function if domain restricted to $[-9,9]$.