To find the domain $D(f)$ and range $R(f)$ of the function $y = \sqrt{2x - 6} - 1$, let's solve step-by-step:

1. Domain $D(f)$:

The domain of the function is determined by the values of $x$ that make the expression under the square root non-negative: $2x - 6 \geq 0$ Solve for $x$: $2x \geq 6 \quad \Rightarrow \quad x \geq 3$ Thus, the domain is: $D(f) = [3, \infty)$

2. Range $R(f)$:

The range is determined by the output values of $y$: $y = \sqrt{2x - 6} - 1$ The square root $\sqrt{2x - 6}$ is always non-negative ($\geq 0$), so: $\sqrt{2x - 6} - 1 \geq -1$ The minimum value of $y$ occurs when $x = 3$: $y = \sqrt{2(3) - 6} - 1 = \sqrt{0} - 1 = -1$ As $x \to \infty$, $\sqrt{2x - 6} \to \infty$, so $y \to \infty$.

Thus, the range is: $R(f) = [-1, \infty)$

Final Answer:

• Domain: $D(f) = [3, \infty)$
• Range: $R(f) = [-1, \infty)$

Would you like a detailed graph to visualize the domain and range, or do you have any questions?

Here are 5 related questions to explore:

1. How do square root functions affect the domain and range of a function?
2. How would the function change if the equation was $y = \sqrt{2x - 6} + 1$?
3. What are the effects of translating a square root function vertically or horizontally?
4. Can we determine the domain and range of $y^2 = 2x - 6$ instead?
5. How would the function behave if $2x - 6 < 0$?

Tip: Always check the square root argument to avoid negative values, as they