Let's break this problem down step by step:

(a) Find the largest possible domains and ranges of $F(x)$ and $G(x)$

1. Domain and Range of $F(x) = \sqrt{-x^2 + 6x - 5}$

• The function $F(x)$ involves a square root, which requires that the expression inside the square root must be non-negative.

$-x^2 + 6x - 5 \geq 0$ Solve the quadratic inequality: $-x^2 + 6x - 5 = 0$ We can solve this using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -1$, $b = 6$, and $c = -5$.

$x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-5)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 20}}{-2} = \frac{-6 \pm \sqrt{16}}{-2}$ $x = \frac{-6 \pm 4}{-2}$ So, $x = \frac{-6 + 4}{-2} = 1$ and $x = \frac{-6 - 4}{-2} = 5$.

Thus, $-x^2 + 6x - 5 \geq 0$ when $1 \leq x \leq 5$.

• Domain of $F(x)$: $[1, 5]$.

• Range of $F(x)$:

  • At $x = 1$, $F(1) = \sqrt{-(1)^2 + 6(1) - 5} = \sqrt{0} = 0$.
  • At $x = 5$, $F(5) = \sqrt{-(5)^2 + 6(5) - 5} = \sqrt{0} = 0$.
  • At $x = 3$, ( F(3) = \sqrt{-(3)^2 + 6(3) - 5} = \sqrt{9 - 18 + 5} = \sqrt