Let’s break this down!

Exercises 3 & 4: Domains and Ranges
3. $f(x) = 2$, $g(x) = x^2 + 1$:

• $f(x) = 2$:

  • Domain: All real numbers ($\mathbb{R}$), as it’s constant.
  • Range: $\{2\}$, since $f(x)$ always equals 2.

• $g(x) = x^2 + 1$:

  • Domain: $\mathbb{R}$ (polynomial function).
  • Range: $[1, \infty)$, since $x^2 \geq 0$.

• $\frac{f}{g} = \frac{2}{x^2 + 1}$:

  • Domain: $\mathbb{R}$, because $g(x) \neq 0$.
  • Range: $(0, 2]$, since $\frac{2}{x^2 + 1} \leq 2$.

4. $f(x) = 1$, $g(x) = 1 + \sqrt{x}$:

   • $f(x) = 1$:

     • Domain: $\mathbb{R}$.
     • Range: $\{1\}$.

   • $g(x) = 1 + \sqrt{x}$:

     • Domain: $[0, \infty)$ (since $\sqrt{x}$ is defined for non-negative $x$).
     • Range: $[1, \infty)$, since $\sqrt{x} \geq 0$.

   • $\frac{f}{g} = \frac{1}{1 + \sqrt{x}}$:

     • Domain: $[0, \infty)$.
     • Range: $(0, 1]$, since $1 + \sqrt{x} \geq 1$.

Exercise 5: Composites
Given $f(x) = x + 5$ and $g(x) = x^2 - 3$:
a. $f(g(0))$:

• $g(0) = 0^2 - 3 = -3$.
• $f(-3) = -3 + 5 = 2$.
• Answer: $2$.

b. $g(f(0))$:

• $f(0) = 0 + 5 = 5$.
• $g(5) = 5^2 - 3 = 25 - 3 = 22$.
• Answer: $22$.

c. $f(g(x))$:

• $g(x) = x^2 - 3$.
• $f(g(x)) = f(x^2 - 3) = (x^2 - 3) + 5 = x^2 + 2$.
• Answer: $x^2 + 2$.

d. $g(f(x))$:

• $f(x) = x + 5$.
• $g(f(x)) = g(x + 5) = (x + 5)^2 - 3 = x^2 + 10x + 25 - 3 = x^2 + 10x + 22$.
• Answer: $x^2 + 10x + 22$.

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5 Related Questions:

1. How do you find the domain of a square root function like $g(x) = 1 + \sqrt{x}$?
2. What is the relationship between composite functions and their domains?
3. How does a constant function’s range differ from that of a polynomial?
4. How do you determine the range of a fraction function like $\frac{f}{g}$?
5. What happens if a composite function results in an undefined value?

Tip: Always verify the domain of inner and outer functions when working with compositions!