Example 9. Find the domain of the function

$$
f(x)=\sqrt{\left(9-x^{2}\right)\left(x^{2}-4\right)}+\sqrt[3]{5 x+7}
$$
Solution. The function is the sum of two functions. The second one is defined for all $x$, since the cube root exists for any $x$. The first function $\sqrt{\left(9-x^{2}\right)\left(x^{2}-4\right)}$ is defined only for those values of $x$ for which the expression under the square root is non-negative (the square root exists only for non-negative numbers). Therefore, it must be

$$
\left(9-x^{2}\right)\left(x^{2}-4\right) \geq 0
$$

This is possible when:

$$
\text { 1) } \left.9-x^{2} \geq 0, x^{2}-4 \geq 0 ; 2\right) 9-x^{2} \leq 0, x^{2}-4 \leq 0
$$

Consider the first condition: $x^{2} \leq 9$, i.e., $|x| \leq 3$ or $-3 \leq x \leq 3$ and $x^{2} \geq 4$, i.e., $|x| \geq 2$ or $x \leq-2$ and $x \geq 2$. Thus, this condition is satisfied when

$$
-3 \leq x \leq-2 \text { and } 2 \leq x \leq 3 .
$$

Now consider the second condition, which can be rewritten as: $x^{2} \geq 9$, i.e., $|x| \geq 3$ or $x \leq-3$ and $x \geq 3$, and $x^{2} \leq 4$, i.e., $|x| \leq 2$ or $-2 \leq x \leq 2$. Therefore, this condition is not feasible (the inequalities $-2 \leq x \leq 2, x3$ cannot be satisfied simultaneously).

Thus, the domain of the function $f_{1}(x)=\sqrt{\left(9-x^{2}\right)\left(x^{2}-4\right)}$ is the union of two segments: $[-3,-2],[2,3]$.

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-011.jpg?height=54&width=403&top_left_y=1279&top_left_x=100)

Fig. 1.1
The sum of two functions is defined where each term is defined, so the original function is also defined in the specified two

segments (Fig. 1.1).

Remark. If a function contains even roots, it is defined only for those values of $x$ for which the expressions under the roots are non-negative.