The given function is:

$f(x) = 3 - 2\sqrt{x - 1}$

Key points:

• This function includes a square root, so we need to ensure the expression under the square root is non-negative to keep the function valid in the real number system.

Domain:

The square root function requires that the argument inside the square root is non-negative, so we must have:

$x - 1 \geq 0 \quad \Rightarrow \quad x \geq 1$

Therefore, the domain of $f(x)$ is:

$\boxed{x \geq 1}$

Range:

As $x \geq 1$, the square root term $\sqrt{x - 1}$ will take values starting from 0 (when $x = 1$) and increase as $x$ increases.

• When $x = 1$, $f(1) = 3 - 2\sqrt{1 - 1} = 3 - 2(0) = 3$.
• As $x \to \infty$, $\sqrt{x - 1}$ grows larger, making $-2\sqrt{x - 1}$ more negative. Hence, $f(x)$ decreases indefinitely as $x \to \infty$.

Thus, the range of $f(x)$ is:

$\boxed{(-\infty, 3]}$

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Related Questions:

1. What is the derivative of $f(x) = 3 - 2\sqrt{x - 1}$?
2. How would you compute the inverse of this function?
3. What is the behavior of the function as $x \to 1^+$?
4. Can you find the second derivative of $f(x)$?
5. How does the function transform if the square root is replaced by a cube root?

Tip:

When dealing with square roots in functions, always check the domain carefully since square roots are only defined for non-negative values in the real number system.