The function given is:

$y = \frac{1 + x}{5 - x}$

Domain:

The domain of a function consists of all values of $x$ for which the function is defined. In this case, the function is undefined when the denominator equals zero. To find when the denominator is zero:

$5 - x = 0$

$x = 5$

Thus, the function is undefined at $x = 5$. Therefore, the domain is all real numbers except $x = 5$. In interval notation, the domain is:

$\text{Domain} = (-\infty, 5) \cup (5, \infty)$

Range:

The range of the function refers to all possible values of $y$. To find the range, let's manipulate the equation and solve for $x$ in terms of $y$:

$y = \frac{1 + x}{5 - x}$

Multiply both sides by $5 - x$:

$y(5 - x) = 1 + x$

Distribute $y$:

$5y - xy = 1 + x$

Group terms involving $x$ on one side:

$5y - 1 = x + xy$

Factor out $x$ on the right-hand side:

$5y - 1 = x(1 + y)$

Now, solve for $x$:

$x = \frac{5y - 1}{1 + y}$

For $x$ to be defined, the denominator $1 + y$ cannot be zero, which gives $y = -1$. Therefore, the function is undefined when $y = -1$.

Thus, the range is all real numbers except $y = -1$. In interval notation, the range is:

$\text{Range} = (-\infty, -1) \cup (-1, \infty)$

Summary:

• Domain: $(-\infty, 5) \cup (5, \infty)$
• Range: $(-\infty, -1) \cup (-1, \infty)$

Would you like further details or have any questions?

Here are some related questions you might want to explore:

1. How can the vertical asymptote of this function be determined?
2. What happens to $y$ as $x \to \infty$?
3. What is the horizontal asymptote of this function?
4. How would you find the intercepts of this function?
5. What is the behavior of this function around $x = 5$?

Tip: The vertical asymptote occurs where the function's denominator is zero.