Key Concepts and Formulas

• Domain of a Rational Function: For a function of the form $f(x) = \frac{P(x)}{Q(x)}$, the denominator $Q(x)$ cannot be zero.
• Domain of a Logarithmic Function: For a function of the form $f(x) = \log_b(g(x))$, the argument $g(x)$ must be strictly positive ($g(x) > 0$), and the base $b$ must be positive and not equal to 1.
• Intersection of Domains: The domain of a composite function is the intersection of the domains of its individual components.

Step-by-Step Solution

The given function is $f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x)$. To find the domain of definition, we need to ensure that both terms are well-defined.

Step 1: Determine the restrictions from the rational term. The first term is $\frac{3}{4 - x^2}$. For this term to be defined, the denominator cannot be zero. $4 - x^2 \neq 0$ Factoring the difference of squares: $(2 - x)(2 + x) \neq 0$ This implies: $2 - x \neq 0 \quad \Rightarrow \quad x \neq 2$ $2 + x \neq 0 \quad \Rightarrow \quad x \neq -2$ So, $x$ cannot be $2$ or $-2$.

Step 2: Determine the restrictions from 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 inequality, we find the roots of the expression $x(x - 1)(x + 1) = 0$, which are $x = -1$, $x = 0$, and $x = 1$. These roots divide the number line into four intervals: $(-\infty, -1)$, $(-1, 0)$, $(0, 1)$, and $(1, \infty)$. We test the sign of the expression 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 positive, so the inequality $x(x - 1)(x + 1) > 0$ holds for $x \in (-1, 0) \cup (1, \infty)$.

Step 3: Combine the restrictions to find the overall domain. The domain of $f(x)$ is the set of all $x$ values that satisfy the conditions from Step 1 and Step 2 simultaneously. From Step 1: $x \neq 2$ and $x \neq -2$. From Step 2: $x \in (-1, 0) \cup (1, \infty)$.

We need to find the intersection of these two conditions. The interval $(-1, 0)$ does not contain $2$ or $-2$. So, this part of the domain remains $(-1, 0)$. The interval $(1, \infty)$ contains $x = 2$. We must exclude $x = 2$ from this interval. This splits the interval $(1, \infty)$ into $(1, 2) \cup (2, \infty)$. The value $x = -2$ is not in $(1, \infty)$, so it does not affect this part.

Combining these results, the domain of $f(x)$ is: $(-1, 0) \cup (1, 2) \cup (2, \infty)$

Common Mistakes & Tips

• Forgetting the "strictly positive" condition for logarithms: The argument of a logarithm must be greater than zero, not greater than or equal to zero.
• Not factoring completely: Always factor polynomial inequalities completely to identify all critical points for sign analysis.
• Confusing union and intersection: The domain of a function composed of multiple parts is the intersection of the domains of each part, meaning all conditions must be met simultaneously.

Summary To find the domain of the function $f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x)$, we first identified the restrictions on $x$ for the rational part (denominator cannot be zero) and the logarithmic part (argument must be strictly positive). The rational part requires $x \neq 2$ and $x \neq -2$. The logarithmic part requires $x \in (-1, 0) \cup (1, \infty)$. The overall domain is the intersection of these conditions, which results in $x \in (-1, 0) \cup (1, 2) \cup (2, \infty)$.

The final answer is \boxed{(-1, 0) \cup (1, 2) \cup (2, \infty)}. This corresponds to option (A).