### Instruction: Find the range of the function
\[j(x) = \frac{5x + 3}{x}.\].### Response: To find the range of a function, I need to consider the possible values of the output, given any input in the domain.
The domain of this function is all real numbers except x = 0, because that would make the denominator zero and the function undefined.
So, for any nonzero x, what values can j(x) take?
I can rewrite the function as
\[j(x) = \frac{5x}{x} + \frac{3}{x} = 5 + \frac{3}{x}.\]
This makes it easier to see how j(x) behaves as x changes.
As x gets very large, either positive or negative, the term 3/x gets very small, approaching zero.
So j(x) gets closer and closer to 5.
As x gets closer to zero from either side, the term 3/x gets very large, either positive or negative.
So j(x) gets farther and farther from 5.
In fact, j(x) can take any value except 5, because there is no value of x that makes 3/x equal to zero.
Therefore, the range of the function is all real numbers except 5.
I can write this using set notation as
\[\{y \in \mathbb{R} : y \neq 5\}.\]