Find the domain and range of the function f(x) = 3/(x-2).

The domain of a function is the set of all possible input values ($x$) for which the function is defined.

• The cube root function $\sqrt[3]{u}$ is defined for all real numbers $u$. So, there are no restrictions from the cube root itself.
• The expression inside the cube root is a fraction $\frac{x-4}{x-2}$. For a fraction to be defined, its denominator cannot be zero. Set the denominator equal to zero to find the excluded value:

$$x-2 = 0$$ $$x = 2$$

So, $x$ cannot be equal to 2. The domain is all real numbers except 2. In interval notation, this is $(-\infty, 2) \cup (2, \infty)$.

The range of a function is the set of all possible output values ($y$) that the function can produce.
Let $y = \sqrt[3]{\frac{x-4}{x-2}}$.
To find the range, we solve for $x$ in terms of $y$.
First, cube both sides of the equation:

$$y^3 = \frac{x-4}{x-2}$$

Multiply both sides by $(x-2)$:

$$y^3(x-2) = x-4$$

Distribute $y^3$:

$$xy^3 - 2y^3 = x-4$$

Move all terms with $x$ to one side and terms without $x$ to the other side:

$$xy^3 - x = 2y^3 - 4$$

Factor out $x$:

$$x(y^3 - 1) = 2y^3 - 4$$

Divide by $(y^3 - 1)$ to solve for $x$:

$$x = \frac{2y^3 - 4}{y^3 - 1}$$

For $x$ to be a real number, the denominator of this expression cannot be zero.
Set the denominator equal to zero to find the excluded value for $y$:

$$y^3 - 1 = 0$$ $$y^3 = 1$$ $$y = \sqrt[3]{1}$$ $$y = 1$$

So, $y$ cannot be equal to 1.
The range is all real numbers except 1.
In interval notation, this is $(-\infty, 1) \cup (1, \infty)$.

The domain and range are:
Domain: $\boxed{x \in (-\infty, 2) \cup (2, \infty)}$
Range: $\boxed{y \in (-\infty, 1) \cup (1, \infty)}$