The problem defines a function $F(x) = 2\sqrt{9-x^2}$ with domain $-3 \le x \le 3$. We need to find the image (value) of $F(2)$ and $F(-2)$, and then find the inverse function $F^{-1}(x)$.

1. Problem Description

The problem defines a function $F(x) = 2\sqrt{9-x^2}$ with domain $-3 \le x \le 3$.

We need to find the image (value) of $F(2)$ and $F(-2)$, and then find the inverse function $F^{-1}(x)$.

2. Solution Steps

(i) Finding the image of 2 and -2:

To find $F(2)$, we substitute $x=2$ into the function:

$F(2) = 2\sqrt{9-(2)^2} = 2\sqrt{9-4} = 2\sqrt{5}$.

To find $F(-2)$, we substitute $x=-2$ into the function:

$F(-2) = 2\sqrt{9-(-2)^2} = 2\sqrt{9-4} = 2\sqrt{5}$.

(ii) Finding the inverse function $F^{-1}(x)$:

Let $y = F(x)$, so $y = 2\sqrt{9-x^2}$.

To find the inverse function, we need to solve for $x$ in terms of $y$:

$y = 2\sqrt{9-x^2}$

$\frac{y}{2} = \sqrt{9-x^2}$

Square both sides:

$(\frac{y}{2})^2 = 9-x^2$

$\frac{y^2}{4} = 9-x^2$

$x^2 = 9 - \frac{y^2}{4}$

$x = \pm\sqrt{9 - \frac{y^2}{4}}$

$x = \pm\sqrt{\frac{36-y^2}{4}}$

$x = \pm\frac{\sqrt{36-y^2}}{2}$

Since the original function had a domain of $-3 \le x \le 3$, and the range of $F(x)$ is $0 \le y \le 6$,

the inverse function $F^{-1}(x)$ must satisfy the domain $0 \le x \le 6$.

Because the original function is an even function, we need to restrict the domain of the original function to find the inverse function.

Assuming $0 \le x \le 3$, we would select $F^{-1}(x) = \frac{\sqrt{36-x^2}}{2}$

If $-3 \le x \le 0$, we select $F^{-1}(x) = -\frac{\sqrt{36-x^2}}{2}$

3. Final Answer

(i) $F(2) = 2\sqrt{5}$, $F(-2) = 2\sqrt{5}$.

(ii) $F^{-1}(x) = \pm \frac{\sqrt{36-x^2}}{2}$ or more specifically, $F^{-1}(x) = \frac{\sqrt{36-x^2}}{2}$ for $0 \le x \le 3$ and $F^{-1}(x) = -\frac{\sqrt{36-x^2}}{2}$ for $-3 \le x \le 0$.