The range of the function f(x) = √(x−1)(3−x) is

The correct option is [1, 3].

To understand why this is the correct choice, let us analyze the conditions under which the given function is defined, which determines its domain (interval of valid values) where the function yields real outputs.

The function is defined as:

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

For the square root to produce a real number, the quantity under the radical must be non-negative. Therefore, we set up the following inequality:

$$(x-1)(3-x)\ge 0$$

To solve this inequality, we can factor out a negative sign from the second term:

$$-(x-1)(x-3)\ge 0$$

Multiplying both sides of the inequality by $-1$ and reversing the direction of the inequality sign yields:

$$(x-1)(x-3)\le 0$$

The boundary points (roots) for this inequality are $x=1$ and $x=3$. For the product to be less than or equal to zero, $x$ must lie between these two values:

$$1\le x\le 3$$

Expressing this in interval notation, we get:

$$x\in [1,3]$$

Thus, the function is defined on the interval [1, 3].