Find the domain of the function $$f(x) = \frac{x + 31}{x + 13}$$ Express you answer in interval notation.
I will assume that the function is a real-valued function of a real variable, meaning that its domain and codomain are subsets of the real numbers.
This is a rational function, meaning that it can be expressed as the quotient of two polynomials. A rational function is undefined if its denominator is equal to zero. Therefore, the domain of a rational function is the set of all real numbers except those that make the denominator equal to zero.
You found that the expression in the denominator is equal to zero when $x = -13$. Therefore, the domain of the function is the set of all real numbers except $-13$. In set notation,
$$\text{Dom}_f = \{x \in \mathbb{R} \mid x \neq 13\}$$
In interval notation, the domain is
$$\text{Dom}_f = (-\infty, -13) \cup (-13, \infty)$$
meaning the union of all real numbers less than $-13$ and all real numbers greater than $-13$.
I found that when making the equations $0$, $x=−31$ and $x=−13$. I can not figure out the brackets.
The quantities $x + 31$ and $x + 13$ are algebraic expressions, not equations. To have an equation, you must set two quantities equal to each other.
By setting the denominator $x + 13$ equal to zero, you determined where the rational function
$$f(x) = \frac{x + 31}{x + 13}$$
is undefined. Determining this tells you which real numbers are not in the domain.
By setting the numerator $x + 31$ equal to zero, you determined where the function is equal to zero and where its graph has an $x$-intercept.
In interval notation, we use a parenthesis when an endpoint is not included. We use a bracket when it is.
$(a, b) = \{x \in \mathbb{R} \mid a < x < b\}$ is the set of all real numbers greater than $a$ and less than $b$.
$[a, b] = \{x \in \mathbb{R} \mid a \leq x \leq b\}$ is the set of all real numbers that are greater than or equal to $a$ and less than or equal to $b$.
$[a, b) = \{x \in \mathbb{R} \mid a \leq x < b\}$ is the set of all real numbers that are greater than or equal to $a$ and less than $b$.
$(a, b] = \{x \in \mathbb{R} \mid a < x \leq b\}$ is the set of all real numbers that are greater than $a$ and less than or equal to $b$.
$(a, \infty) = \{x \in \mathbb{R} \mid x > a\}$ is the set of all real numbers greater than $a$.
$[a, \infty) = \{x \in \mathbb{R} \mid x \geq a\}$ is the set of all real numbers that are greater than or equal to $a$.
$(-\infty, b) = \{x \in \mathbb{R} \mid x < b\}$ is the set of all real numbers less than $b$.
$(-\infty, b] = \{x \in \mathbb{R} \mid x \leq b\}$ is the set of all real numbers less than or equal to $b$.
$(-\infty, \infty) = \mathbb{R}$ is the set of all real numbers.
Find the domain of the function $$f(x) = \frac{38x}{x^2 + 4x - 5}$$ Express your answer in interval notation.
As stated above, the domain of the rational function is the set of all real numbers except those that make the denominator equal to zero. We solve the equation $x^2 + 4x - 5 = 0$ by factoring.
\begin{align*}
x^2 + 4x - 5 & = 0\\
(x + 5)(x - 1) & = 0
\end{align*}
Setting each factor equal to zero yields
\begin{align*}
x + 5 & = 0 & x - 1 & = 0\\
x & = -5 & x & = 1
\end{align*}
Hence, the domain is the set of all real numbers except $-5$ and $1$. In set notation,
$$\text{Dom}_f = \{x \in \mathbb{R} \mid x \neq -5, 1\}$$
In interval notation,
$$\text{Dom}_f = (-\infty, -5) \cup (-5, 1) \cup (1, \infty)$$
Best Answer
The domain of $f\circ g$ is $\mathbb{R}\setminus\{-11,-5\}=(-\infty,-11)\cup(-11,-5)\cup(-5,+\infty)$. That's because: