Find the domain of the function $f(x) = \frac{\sqrt{x-2}}{\sqrt{5}} + \sqrt{(x - 1)(x + 2)(x + 4)}.
To find the domain of a function, we need to consider values of $x$ that make the function undefined.
For the first term, $\frac{\sqrt{x - 2}}{\sqrt{5}}$, the function is undefined when $x - 2 < 0$, which means $x < 2$. However, there are no restrictions on the numerator because we are taking the square root of $x - 2$, which is always nonnegative. So the domain of the first term is $x \ge 2$.
For the second term, $\sqrt{(x - 1)(x + 2)(x + 4)}$, the function is undefined when the expression inside the square root becomes negative. This occurs when the product $(x - 1)(x + 2)(x + 4)$ is negative. Since we are taking the square root of this product, the expression must be nonnegative. This means we need to find when the product is zero.
The term $x - 1$ is equal to 0 when $x = 1$, the term $x + 2$ is equal to 0 when $x = -2$, and the term $x + 4$ is equal to 0 when $x = -4$. So the product $(x - 1)(x + 2)(x + 4)$ is equal to 0 when $x = -4$, $x = -2$, or $x = 1$.
Therefore, the domain of the second term is $x \neq -4, -2, 1$. In other words, $x$ can be any real number except $x = -4, -2,$ or $1$.
Since the domain of $f(x)$ is the intersection of the domains of each term, we have $x \ge 2$ and $x \neq -4, -2, 1$. Therefore, the domain of $f(x)$ is $x \in \boxed{[2, \infty)} \setminus \{-4, -2, 1\}$.
To find the domain of the function, we need to determine the values of $x$ for which the function is defined.
First, let's consider the denominator of the first term, $\sqrt{5}$. Since the square root of $5$ is always positive, the denominator does not affect the domain.
Next, we need to consider the radicand (the expression inside the square root function) of the first term, $x-2$. For the square root function to be defined, the radicand must be equal to or greater than $0$. Therefore, we have:
$x - 2 \geq 0$
Solving this inequality, we find:
$x \geq 2$
Thus, the first term is defined for all values of $x$ greater than or equal to $2$.
Moving on to the second term, we have the expression $\sqrt{(x - 1)(x + 2)(x + 4)}$. Again, the square root function is defined when the radicand is greater than or equal to $0$. So, we need to solve the inequality:
$(x - 1)(x + 2)(x + 4) \geq 0$
To solve this inequality, we can use a sign chart or plot the critical points on a number line.
The critical points occur when the factors $(x - 1)$, $(x + 2)$, and $(x + 4)$ are equal to $0$. These are the values of $x$ that make the inequality equal to $0$. Therefore, we have:
$x - 1 = 0 \implies x = 1$
$x + 2 = 0 \implies x = -2$
$x + 4 = 0 \implies x = -4$
Plotting these critical points on a number line:
-------------------◉-----------------◉----------------------◉--------------------
-4 -2 1
We choose a test point in each of the three intervals defined by the critical points to determine the sign of the expression $(x - 1)(x + 2)(x + 4)$ in that interval.
For $x < -4$, let's choose $x = -5$. Plugging this value into the expression, we have:
$(-5 - 1)(-5 + 2)(-5 + 4) = (-6)(-3)(-1) = 18$
Since this is positive, the expression is positive in the interval $(-\infty, -4)$.
For $-4 < x < -2$, let's choose $x = -3$. Plugging this value into the expression, we have:
$(-3 - 1)(-3 + 2)(-3 + 4) = (-4)(-1)(1) = 4$
Since this is positive, the expression is positive in the interval $(-4, -2)$.
For $-2 < x < 1$, let's choose $x = 0$. Plugging this value into the expression, we have:
$(0 - 1)(0 + 2)(0 + 4) = (-1)(2)(4) = -8$
Since this is negative, the expression is negative in the interval $(-2, 1)$.
For $x > 1$, let's choose $x = 2$. Plugging this value into the expression, we have:
$(2 - 1)(2 + 2)(2 + 4) = (1)(4)(6) = 24$
Since this is positive, the expression is positive in the interval $(1, \infty)$.
Therefore, the inequality $(x - 1)(x + 2)(x + 4) \geq 0$ is satisfied when $x$ is less than or equal to $-4$, or between $-2$ and $1$, or greater than or equal to $1$.
Combining this with the first term, the function is defined when $x$ is greater than or equal to $2$ and less than or equal to $-4$, or between $-2$ and $1$, or greater than or equal to $1$.
So, the domain of the function $f(x)$ is given by:
$D = (-\infty, -4] \cup (-2, 1) \cup [2,\infty)$