Questions: Let f(x)=x^3+1 and g(x)=sqrt(x^2-16). Determine the domain of f/g(x).

Transcript text: Let $f(x)=x^{3}+1$ and $g(x)=\sqrt{x^{2}-16}$. Determine the domain of $\frac{f}{g}(x)$.

Solution

Solution Steps

To determine the domain of $\frac{f}{g}(x)$, we need to find the values of $x$ for which $g(x) \neq 0$ and $g(x)$ is defined. Since $g(x) = \sqrt{x^2 - 16}$, it is defined when $x^2 - 16 \geq 0$. Additionally, $g(x) \neq 0$ implies $x^2 - 16 \neq 0$. Solve these inequalities to find the domain of $\frac{f}{g}(x)$.

Step 1: Determine the Domain of $ f(x) $

The function $ f(x) = x^3 + 1 $ is a polynomial function. Polynomial functions are defined for all real numbers. Therefore, the domain of $ f(x) $ is:

\[ \text{Domain of } f(x): \quad (-\infty, \infty) \]

Step 2: Determine the Domain of $ g(x) $

The function $ g(x) = \sqrt{x^2 - 16} $ is a square root function. The expression inside the square root, $ x^2 - 16 $, must be non-negative for $ g(x) $ to be defined. Therefore, we solve the inequality:

\[ x^2 - 16 \geq 0 \]

This can be factored as:

\[ (x - 4)(x + 4) \geq 0 \]

To solve this inequality, we find the critical points where the expression equals zero:

\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

We test intervals around these critical points:

$ x < -4 $
$ -4 \leq x \leq 4 $
$ x > 4 $

For $ x < -4 $, choose $ x = -5 $: \[ (-5 - 4)(-5 + 4) = (-9)(-1) = 9 \geq 0 \]
For $ -4 \leq x \leq 4 $, choose $ x = 0 $: \[ (0 - 4)(0 + 4) = (-4)(4) = -16 < 0 \]
For $ x > 4 $, choose $ x = 5 $: \[ (5 - 4)(5 + 4) = (1)(9) = 9 \geq 0 \]

Thus, the solution to the inequality is:

\[ x \in (-\infty, -4] \cup [4, \infty) \]

Step 3: Determine the Domain of $ \frac{f}{g}(x) $

The function $ \frac{f}{g}(x) = \frac{x^3 + 1}{\sqrt{x^2 - 16}} $ is defined where both $ f(x) $ and $ g(x) $ are defined, and $ g(x) \neq 0 $. Since $ g(x) $ is defined for $ x \in (-\infty, -4] \cup [4, \infty) $, and $ g(x) \neq 0 $ for these intervals, the domain of $ \frac{f}{g}(x) $ is the same as the domain of $ g(x) $.