Key Concepts and Formulas

• Domain of a Rational Function: The denominator of a rational function cannot be zero. If $f(x) = \frac{N(x)}{D(x)}$, then $D(x) \neq 0$.
• Domain of a Logarithmic Function: The argument of a logarithm must be strictly positive. If $f(x) = \log_b(g(x))$, then $g(x) > 0$.
• Domain of a Sum of Functions: The domain of a sum of functions is the intersection of the domains of each individual function.
• Solving Polynomial Inequalities: Use the wavy curve method (sign analysis) by finding the roots of the polynomial and testing the sign in the intervals defined by these roots.

Step-by-Step Solution

We are given the function $f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$. To find the domain of $f(x)$, we need to ensure that both terms are defined simultaneously.

Step 1: Determine the domain of the rational term. The first term is ${1 \over {4 - {x^2}}}$. For this term to be defined, the denominator cannot be zero. $4 - x^2 \neq 0$ Factoring the difference of squares, we get: $(2 - x)(2 + x) \neq 0$ This implies: $2 - x \neq 0 \quad \text{and} \quad 2 + x \neq 0$ $x \neq 2 \quad \text{and} \quad x \neq -2$ So, the domain for the first term is $x \in \mathbb{R} \setminus \{-2, 2\}$.

Step 2: Determine the domain of the logarithmic term. The second term is ${\log _{10}}({x^3} - x)$. For this term to be defined, the argument of the logarithm must be strictly positive. $x^3 - x > 0$ Factor out $x$: $x(x^2 - 1) > 0$ Factor the difference of squares $(x^2 - 1)$: $x(x - 1)(x + 1) > 0$ To solve this polynomial inequality, we find the roots of the expression $x(x-1)(x+1)$, which are $x = -1, 0, 1$. These roots divide the number line into four intervals: $(-\infty, -1)$, $(-1, 0)$, $(0, 1)$, and $(1, \infty)$. We analyze the sign of $x(x-1)(x+1)$ in each interval.

• For $x < -1$ (e.g., $x = -2$): $(-2)(-2-1)(-2+1) = (-2)(-3)(-1) = -6 < 0$.
• For $-1 < x < 0$ (e.g., $x = -0.5$): $(-0.5)(-0.5-1)(-0.5+1) = (-0.5)(-1.5)(0.5) = 0.375 > 0$.
• For $0 < x < 1$ (e.g., $x = 0.5$): $(0.5)(0.5-1)(0.5+1) = (0.5)(-0.5)(1.5) = -0.375 < 0$.
• For $x > 1$ (e.g., $x = 2$): $(2)(2-1)(2+1) = (2)(1)(3) = 6 > 0$.

We need the expression to be strictly greater than zero, so the domain for the logarithmic term is $x \in (-1, 0) \cup (1, \infty)$.

Step 3: Find the intersection of the domains of both terms. The domain of $f(x)$ is the set of all $x$ values that satisfy both conditions from Step 1 and Step 2. We need to find the intersection of: Domain 1: $x \in (-\infty, -2) \cup (-2, 2) \cup (2, \infty)$ Domain 2: $x \in (-1, 0) \cup (1, \infty)$

Let's visualize this on a number line. The intervals from Domain 2 are $(-1, 0)$ and $(1, \infty)$. We need to check which parts of these intervals are excluded by Domain 1. Domain 1 excludes $x = -2$ and $x = 2$.

• Consider the interval $(-1, 0)$ from Domain 2. This interval does not contain $-2$ or $2$. So, $(-1, 0)$ is part of the intersection.
• Consider the interval $(1, \infty)$ from Domain 2. This interval contains $x = 2$, which is excluded by Domain 1. Therefore, we must exclude $x=2$ from $(1, \infty)$. This gives us $(1, 2) \cup (2, \infty)$.

Combining the valid intervals from both domains, the intersection is: $(-1, 0) \cup (1, 2) \cup (2, \infty)$

Common Mistakes & Tips

• Forgetting the strict inequality for logarithms: Remember that the argument of a logarithm must be strictly positive ($>0$), not non-negative ($\geq 0$).
• Errors in sign analysis: Carefully test values in each interval when solving polynomial inequalities. A single incorrect sign can lead to the wrong interval.
• Missing the intersection: The domain of the entire function is the intersection of the domains of its constituent parts, not the union.

Summary To find the domain of the given function, we first determined the restrictions imposed by the rational part (denominator not equal to zero) and the logarithmic part (argument strictly positive). The domain of the function is the set of all real numbers where both these conditions are met simultaneously. By finding the intersection of these individual domains, we arrived at the final domain of the function.

The final answer is \boxed{(-1, 0) \cup (1, 2) \cup (2, \infty)}.