Question

(e) $h(x) = \frac{x - 4}{x^2 + 9}$
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 $\vee$ 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:
$x^2 + 9 \geq 0 + 9$
$x^2 + 9 \geq 9$

Step2: Compare to zero

Since $x^2 + 9 \geq 9$, it is always greater than 0 for all real $x$.

Step3: Determine domain

No real $x$ makes the denominator zero, so domain is all real numbers.

Answer:

Select: $\boldsymbol{>}$
Domain: All real numbers (or $\boldsymbol{(-\infty, \infty)}$)