what is the maximum value of the function (see right) over the interval -3 < x < 3? f(x) = 18 / (x² - 9) 0 2…

Answer:

D. -2

Explanation:

Step1: Analyze the function

The function is $f(x)=\frac{18}{x^{2}-9}=\frac{18}{(x - 3)(x + 3)}$, and the interval is $-3<x<3$.

Step2: Consider the behavior of the denominator

In the interval $-3<x<3$, $x^{2}-9<0$. As $x$ approaches $- 3$ from the right or $3$ from the left, the denominator $x^{2}-9$ approaches $0$ from the negative - side.

Step3: Rewrite the function for analysis

Let's consider the function in the given interval. We can rewrite $y = f(x)=\frac{18}{x^{2}-9}$. When $x = 0$, $y=\frac{18}{0 - 9}=-2$. As $x$ moves closer to $-3$ or $3$ within the open - interval $(-3,3)$, the value of $y$ approaches negative infinity. So the maximum value of the function $f(x)$ in the interval $-3<x<3$ is $-2$.