(a) Finding the Largest Possible Domains and Ranges

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

• The square root function requires that the expression inside the square root must be non-negative: $-x^2 + 6x - 5 \geq 0$

  First, solve the quadratic inequality: $-x^2 + 6x - 5 = 0$ Multiplying through by $-1$: $x^2 - 6x + 5 = 0$ Factor the quadratic: $(x - 5)(x - 1) = 0$ So the solutions to the equation are $x = 1$ and $x = 5$. These divide the number line into three intervals: $(-\infty, 1)$, $(1, 5)$, and $(5, \infty)$.

  Test each interval to determine where the quadratic is non-negative:

  • For $x = 0$ (in $(-\infty, 1)$): $-x^2 + 6x - 5 = -5 < 0$
  • For $x = 3$ (in $(1, 5)$): $-x^2 + 6x - 5 = 4 \geq 0$
  • For $x = 6$ (in $(5, \infty)$): $-x^2 + 6x - 5 = -11 < 0$

  Therefore, the domain of $F(x)$ is: $\text{Domain of } F(x) = [1, 5]$

Range of $F(x)$

The range of a square root function is always non-negative. At $x = 1$, we get: $F(1) = \sqrt{-(1)^2 + 6(1) - 5} = \sqrt{0} = 0$ At $x = 5$, we get: $F(5) = \sqrt{-(5)^2 + 6(5) - 5} = \sqrt{0} = 0$ The maximum value occurs at the midpoint of the domain $x = 3$: $F(3) = \sqrt{-(3)^2 + 6(3) - 5} = \sqrt{4} = 2$ Thus, the range of $F(x)$ is: $\text{Range of } F(x) = [0, 2]$

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2. Domain of $G(x) = \frac{3}{x^3 - 3x^2 + 3x - 1}$

The denominator $x^3 - 3x^2 + 3x - 1$ must not be zero. Factor the cubic: $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$ So the function is undefined when $x = 1$. Therefore, the domain of $G(x)$ is: $\text{Domain of } G(x) = (-\infty, 1) \cup (1, \infty)$

Range of $G(x)$

As a rational function with a cubic denominator and a constant numerator, $G(x)$ can take on any real number except where the denominator is zero. Therefore, the range is: $\text{Range of } G(x) = \mathbb{R} \setminus \{0\}$

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(b) Finding $(F \circ G)(x)$ and Its Largest Possible Domain

1. Expression for $(F \circ G)(x)$

The composition $(F \circ G)(x)$ means $F(G(x))$, so: $(F \circ G)(x) = F\left( \frac{3}{(x-1)^3} \right) = \sqrt{-\left( \frac{3}{(x-1)^3} \right)^2 + 6\left( \frac{3}{(x-1)^3} \right) - 5}$ This expression is quite complex, but the domain is determined by the following conditions:

• $G(x)$ must be in the domain of $F(x)$, which is $[1, 5]$.
• $G(x) = \frac{3}{(x - 1)^3}$ must satisfy $1 \leq G(x) \leq 5$.

2. Solving for Domain of $(F \circ G)(x)$

Solve for when $G(x) \in [1, 5]$: $1 \leq \frac{3}{(x-1)^3} \leq 5$ This involves solving inequalities, but let's outline the result:

The largest possible domain of $(F \circ G)(x)$ is found by checking values that satisfy both inequalities. The solution set would be within the intervals where $G(x) \in [1, 5]$.

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Would you like more details on any part of the solution?
Here are 5 related questions you might find interesting:

1. How can I solve inequalities involving rational functions more easily?
2. What are techniques to determine the range of complex functions?
3. Can you explain how to factor higher-degree polynomials efficiently?
4. How do we approach quadratic inequalities like the one for $F(x)$?
5. What makes the composition of functions more restrictive for domains?

Tip: Always check the domain constraints when composing functions—composition can introduce new restrictions!