If $f(x)$ is a function whose domain is $[-8,8]$, and $g(x)=f\left((x^2 - 2)/(x + 1)\right)$, then the domain of $g(x)$ is an interval of what width?
Recall that the domain of a function consists of all the values of $x$ for which the function is defined. In this case, $g(x)$ is defined only when $f\left(\frac{x^2 - 2}{x + 1}\right)$ is defined. Thus we need to consider when $\frac{x^2 - 2}{x + 1}$ is in the domain of $f(x)$. Notice that $f(x)$ is defined for all values of $x$ in the domain $[-8,8]$. Therefore, $\frac{x^2 - 2}{x + 1}$ is always defined except when $x=-1$. Thus the domain of $g(x)$ is an interval of length $\boxed{2}$ because it includes $\frac{x^2 - 2}{x + 1}$ for all $x$ except $x=-1$.
To find the domain of $g(x)$, we need to determine the values of $x$ for which the expression $\frac{x^2 - 2}{x + 1}$ is defined.
First, let's look at the denominator $x + 1$. The expression is defined as long as the denominator is not equal to zero. Therefore, $x + 1 \neq 0$. Solving this equation, we have $x \neq -1$.
Now, let's consider the numerator $x^2 - 2$. There are no restrictions on the domain of the numerator, so we can ignore it.
Combining the restrictions on the numerator and denominator, we find that the domain of $\frac{x^2 - 2}{x + 1}$ is the set of all real numbers except $x = -1$.
Since $f(x)$ has a domain of $[-8, 8]$, the values of $x$ that result from $\frac{x^2 - 2}{x + 1}$ should also be within the domain of $f(x)$. Therefore, the domain of $g(x)$ is the interval $[-8, 8]$ excluding $x = -1$.
The width of this interval is $8 - (-8) = 16$. Thus, the domain of $g(x)$ is an interval of width 16.
To determine the domain of $g(x)$, we need to analyze the given function $g(x)=f\left(\frac{x^2 - 2}{x + 1}\right)$.
First, let's consider the domain of $f(x)$. We are told that the domain of $f(x)$ is $[-8,8]$, which means that $f(x)$ is defined for all values of $x$ in the interval $[-8,8]$.
Next, we look at the expression $\frac{x^2 - 2}{x + 1}$ inside the function $g(x)$. We need to determine the valid values of $x$ that make this expression well-defined.
Since division by zero is undefined, we know that $x + 1 \neq 0$. Solving for $x$, we find that $x \neq -1$.
Now, let's consider the term $x^2 - 2$. The domain of this term is all real numbers since there are no restrictions on $x$ for this term.
Combining these two conditions, we find that the domain of $g(x)$ is given by all values of $x$ that satisfy both conditions: $x \neq -1$ and $x$ belongs to the domain of $f(x)$, which is $[-8,8]$.
Therefore, the interval of the domain of $g(x)$ is $[-8, -1) \cup (-1, 8]$, which spans a width of $[-8, 8]$ minus the point $-1$. Consequently, the width of the domain interval is $8 - (-8) - 1$, which simplifies to $16 - 1 = 15$. Thus, the width of the domain of $g(x)$ is 15 units.