Let F(x) = [−x^(2) + 6x − 5]^(1/2) and G(x) = 3/[x^(3)-3x^(2)+3x-1].


(a) Find their largest possible domains and ranges.(b) Find (F ◦ G)(x) and states its largest possible domain

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

Let F(x) = [−x^2 + 6x − 5]^(1/2) and G(x) = 3/[x^3-3x^2+3x-1]. (a) Find their largest possible domains and ranges. (b) Find (F ◦ G)(x) and state its largest possible domain.

Solve for the largest possible domains and ranges of the functions F(x) = √(−x^2 + 6x − 5) and G(x) = 3/[x^3-3x^2+3x-1]. Learn how to find the composition (F ◦ G)(x) and determine its largest possible domain. This problem involves quadratic equations, square root functions, and function composition. Suitable for Grades 10-12.

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