To determine the domain and range of $g(x)$, we will analyze the given definitions of $f(x)$ and $g(x)$:

1. Definition of $f(x)$: $f(x) = \sqrt{x}, \quad x \in [0, 4]$.

   • Domain of $f(x)$: $[0, 4]$ (since square roots require $x \geq 0$ and the interval is restricted to $[0, 4]$).
   • Range of $f(x)$: $[0, 2]$ (since $\sqrt{4} = 2$).

2. Definition of $g(x)$: $g(x) = -2f(2x - 4) + 3$.

   • For $f(2x - 4)$ to be defined, we require: $2x - 4 \in [0, 4].$ Solving this inequality: $0 \leq 2x - 4 \leq 4 \implies 4 \leq 2x \leq 8 \implies 2 \leq x \leq 4.$ Therefore, the domain of $g(x)$ is $[2, 4]$.

   • For the range of $g(x)$:

     • When $x \in [2, 4]$, $2x - 4$ maps to $[0, 4]$, which is the domain of $f(x)$.
     • The output of $f(x)$ is $[0, 2]$, as established earlier.
     • Substituting into $g(x) = -2f(2x - 4) + 3$, we transform the range: $y = -2f(2x - 4) + 3, \quad f(2x - 4) \in [0, 2].$ Let $z = f(2x - 4)$, where $z \in [0, 2]$: $g(x) = -2z + 3.$ For $z = 0$, $g(x) = -2(0) + 3 = 3$. For $z = 2$, $g(x) = -2(2) + 3 = -4 + 3 = -1$. Therefore, the range of $g(x)$ is $[-1, 3]$.

Final Answer:

• Domain of $g(x)$: $[2, 4]$.
• Range of $g(x)$: $[-1, 3]$.

Would you like further details or help with visualizing this function? Here are some related questions to deepen your understanding:

1. How can we graph $g(x)$ to confirm its domain and range?
2. What transformations were applied to $f(x)$ to create $g(x)$?
3. How do we verify the range using algebraic bounds?
4. Can we find any critical points of $g(x)$ within its domain?
5. What would happen to the domain and range if $f(x)$ were defined on a different interval?

Tip: Always start by analyzing the inner function's domain before applying transformations to ensure accurate results.