Express the domain of f(x) = x / (cube root of (x^2 - 36)) in interval notation.

Step-by-step solution

1. 1

   Decide what the cube root restricts. Odd roots are defined for every real input — $\sqrt[3]{-8} = -2$ is perfectly legitimate. So, unlike a square root, $\sqrt[3]{x^2-36}$ imposes no sign condition on $x^2 - 36$. Treating it like a square root and demanding $x^2 \ge 36$ is the classic error here.

2. 2

   Find where the denominator vanishes. The only genuine restriction is division by zero, and $\sqrt[3]{A} = 0$ exactly when $A = 0$:

   $x^2 - 36 = 0 \quad\Longrightarrow\quad x = \pm 6$

3. 3

   Exclude those two points. Removing $-6$ and $6$ from $\mathbb{R}$ leaves

   $(-\infty,-6) \cup (-6,6) \cup (6,\infty)$

4. 4

   Confirm a value where a square root would fail. At $x = 0$ the radicand is $-36$, and $\sqrt[3]{-36} = -3.301927$, so $f(0) = \tfrac{0}{-3.301927} = 0$ — defined. A square-root version of this function would have excluded the whole interval $(-6,6)$.

5. 5

   Check the behaviour near the excluded points. As $x \to 6^{-}$, $x^2 - 36 \to 0^{-}$ and the cube root tends to $0$ from below, so $f(x) \to -\infty$; from the right it tends to $+\infty$. Both are genuine vertical asymptotes, confirming $x = \pm6$ must be excluded rather than patched.