Question

(c) $h(x) = \frac{x - 4}{x^2 + 9}$
the domain is all real numbers except for those that make the denominator zero.
the variable $x$ has the restriction that $x^2 + 9 \
eq 0$. the expression $x^2 \geq 0$ for all real numbers $x$.
therefore, $x^2 + 9$ select $0$ for all real numbers $x$.
domain: \square

Explanation:

Step1: Analyze denominator inequality

We know $x^2 \geq 0$ for all real $x$. Add 9 to both sides:
$x^2 + 9 \geq 0 + 9$
$x^2 + 9 \geq 9$
Since $9 > 0$, this means $x^2 + 9 > 0$ for all real $x$.

Step2: Determine domain

The denominator can never equal 0, so there are no restrictions on real $x$.

Answer:

Select: $>$
Domain: All real numbers (or in interval notation, $(-\infty, \infty)$)