Let's analyze the problem.

We are given three functions:

• $f(x) = x^2 - 3x + 3$
• $g(x) = \frac{4}{\sqrt{x}}$
• $h(x) = \sqrt{1 - x}$

We are tasked with evaluating:

1. $f(-6)$
2. $g(-6)$
3. $h(-6)$
4. $h(d)$
5. $h(-x)$

Step-by-step solutions:

1. $f(-6)$

Substitute $x = -6$ into $f(x)$: $f(-6) = (-6)^2 - 3(-6) + 3 = 36 + 18 + 3 = 57$ Answer: $f(-6) = 57$.

2. $g(-6)$

Substitute $x = -6$ into $g(x) = \frac{4}{\sqrt{x}}$: $g(-6) = \frac{4}{\sqrt{-6}}$ Since the square root of a negative number is not a real number, the solution is not a real number.

Answer: $g(-6) = \text{not a real number}$.

3. $h(-6)$

Substitute $x = -6$ into $h(x) = \sqrt{1 - x}$: $h(-6) = \sqrt{1 - (-6)} = \sqrt{1 + 6} = \sqrt{7}$ Answer: $h(-6) = \sqrt{7}$.

4. $h(d)$

For $h(d)$, substitute $x = d$ into $h(x) = \sqrt{1 - x}$: $h(d) = \sqrt{1 - d}$ Answer: $h(d) = \sqrt{1 - d}$.

5. $h(-x)$

For $h(-x)$, substitute $x = -x$ into $h(x) = \sqrt{1 - x}$: $h(-x) = \sqrt{1 - (-x)} = \sqrt{1 + x}$ Answer: $h(-x) = \sqrt{1 + x}$.

Final Answers:

1. $f(-6) = 57$
2. $g(-6) = \text{not a real number}$
3. $h(-6) = \sqrt{7}$
4. $h(d) = \sqrt{1 - d}$
5. $h(-x) = \sqrt{1 + x}$

Would you like further clarifications or visual representations?

Related Questions:

1. How do you determine if a solution is not a real number?
2. What are the conditions for a square root to yield a real number?
3. Can $f(x)$ ever yield complex solutions with different inputs?
4. How would the value of $h(x)$ change if $x > 1$?
5. Can $g(x)$ ever yield a real value for negative inputs?

Tip:

Always verify the domain of square roots and denominators to determine whether a function has real or complex outputs.