Question: Consider the function $y = \sqrt{4 - x}$. What are the domain and range of this function? domain: $x \leq 2$, range: $y \geq 2$ domain: $x \leq 4$, range: $y \geq 0$ domain: $x \geq 4$, range: $y \geq 0$ domain: $x \geq 2$, range: $y \geq 2$
1. **State the problem:** We need to find the domain and range of the function $$y = \sqrt{4 - x}$$.
2. **Recall the domain rule for square roots:** The expression inside the square root must be greater than or equal to zero because the square root of a negative number is not a real number.
3. **Find the domain:** Set the inside of the root $$4 - x \geq 0$$.
4. Solve the inequality:
$$4 - x \geq 0$$
$$-x \geq -4$$
Dividing both sides by $$-1$$ reverses the inequality:
$$\cancel{-}x \geq \cancel{-}4 \Rightarrow x \leq 4$$
5. **Domain:** $$x \leq 4$$.
6. **Find the range:** Since $$y = \sqrt{4 - x}$$, the square root function outputs values $$y \geq 0$$.
7. The smallest value inside the root is when $$x = 4$$, giving $$y = \sqrt{0} = 0$$.
8. The largest value inside the root is when $$x$$ is as small as possible (approaching $$-\infty$$), but since the domain restricts $$x \leq 4$$, the maximum inside the root is at $$x = -\infty$$, but practically the function decreases as $$x$$ increases.
9. So the range is $$y \geq 0$$.
**Final answer:**
- Domain: $$x \leq 4$$
- Range: $$y \geq 0$$