Answer :
To determine the domain of the function \( h(x) = \frac{9x}{x(x^2 - 36)} \), we need to find where the function is undefined. The function is undefined wherever the denominator is zero, as division by zero is not possible.
First, let's rewrite the denominator to find its roots:
[tex]\[ x(x^2 - 36) \][/tex]
We notice that the term \( x^2 - 36 \) can be factored further:
[tex]\[ x(x^2 - 36) = x(x - 6)(x + 6) \][/tex]
The function is undefined when any factor of the denominator equals zero. We solve each factor set equal to zero:
1. \( x = 0 \)
2. \( x - 6 = 0 \implies x = 6 \)
3. \( x + 6 = 0 \implies x = -6 \)
So, the function is undefined at \( x = 0 \), \( x = 6 \), and \( x = -6 \).
Therefore, the domain of the function \( h(x) \) includes all real numbers except \( x = 0 \), \( x = 6 \), and \( x = -6 \).
This corresponds to the option:
[tex]\[ B. \{x \mid x \neq \pm 6, x \neq 0\} \][/tex]
Thus, the best answer is:
[tex]\[ \boxed{B} \][/tex]
First, let's rewrite the denominator to find its roots:
[tex]\[ x(x^2 - 36) \][/tex]
We notice that the term \( x^2 - 36 \) can be factored further:
[tex]\[ x(x^2 - 36) = x(x - 6)(x + 6) \][/tex]
The function is undefined when any factor of the denominator equals zero. We solve each factor set equal to zero:
1. \( x = 0 \)
2. \( x - 6 = 0 \implies x = 6 \)
3. \( x + 6 = 0 \implies x = -6 \)
So, the function is undefined at \( x = 0 \), \( x = 6 \), and \( x = -6 \).
Therefore, the domain of the function \( h(x) \) includes all real numbers except \( x = 0 \), \( x = 6 \), and \( x = -6 \).
This corresponds to the option:
[tex]\[ B. \{x \mid x \neq \pm 6, x \neq 0\} \][/tex]
Thus, the best answer is:
[tex]\[ \boxed{B} \][/tex]
