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The problem here requires us to find the values of \( a \) and \( b \) to make the function \( f(x) \) continuous on the interval \( [0, \infty) \).

The function \( f(x) \) is defined as:
\[
f(x) = \begin{cases} 
\frac{a x^3 + b}{x^2 - 4}, & 0 \leq x < 2, \\
x^3 - 3x + 1, & x \geq 2 
\end{cases}
\]

For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit as \( x \) approaches 2 must be equal to the right-hand limit as \( x \) approaches 2, and both must be equal to \( f(2) \).

### Step-by-Step Solution:

1. **Evaluate \( f(2) \):**
   \[
   f(2) = 2^3 - 3 \cdot 2 + 1 = 8 - 6 + 1 = 3.
   \]

   Therefore, \( f(x) \) must approach 3 from both sides as \( x \to 2 \).

2. **Left-Hand Limit as \( x \to 2^- \):**
   We have:
   \[
   \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{a x^3 + b}{x^2 - 4}.
   \]

   Notice that \( x^2 - 4 = (x - 2)(x + 2) \). When \( x \to 2 \), the denominator approaches zero, so we need the numerator to also approach zero for the limit to be finite. Set up the condition:
   \[
   a \cdot 2^3 + b = 0.
   \]
   This gives:
   \[
   8a + b = 0 \quad \Rightarrow \quad b = -8a.
   \]

3. **Substitute \( b = -8a \) and Calculate the Limit:**
   Substitute \( b = -8a \) into the left-hand limit:
   \[
   \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{a x^3 - 8a}{(x - 2)(x + 2)}.
   \]

   Factor the numerator:
   \[
   = \lim_{x \to 2^-} \frac{a (x^3 - 8)}{(x - 2)(x + 2)} = \lim_{x \to 2^-} \frac{a (x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}.
   \]

   Cancel \( x - 2 \) from the numerator and denominator:
   \[
   = \lim_{x \to 2^-} \frac{a (x^2 + 2x + 4)}{x + 2}.
   \]

   Substitute \( x = 2 \):
   \[
   = \frac{a (2^2 + 2 \cdot 2 + 4)}{2 + 2} = \frac{a (4 + 4 + 4)}{4} = \frac{12a}{4} = 3a.
   \]

   For continuity, this limit must equal \( f(2) = 3 \):
   \[
   3a = 3 \quad \Rightarrow \quad a = 1.
   \]

4. **Find \( b \):**
   Substitute \( a = 1 \) into \( b = -8a \):
   \[
   b = -8 \cdot 1 = -8.
   \]

### Final Answer:
The values of \( a \) and \( b \) that make \( f(x) \) continuous on \( [0, \infty) \) are:
\[
a = 1, \quad b = -8.
\]

Would you like further details or have any questions about this solution?

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Here are some related questions to explore further:

1. What is the definition of continuity for piecewise functions?
2. How do we handle removable discontinuities in rational functions?
3. What is the significance of factoring when evaluating limits involving zero denominators?
4. How would the solution change if the piecewise condition had a different breakpoint?
5. How does the factorization of polynomials help simplify limits?

**Tip:** Always check both the left-hand and right-hand limits for continuity, especially at points where the function's definition changes.

Tentukan nilai a dan b agar f(x) = \left(\frac{ax^3 + b}{x^2 - 4}, \; 0 \leq x < 2 \; \text{and} \; x^3 - 3x + 1, \; x \geq 2 \right) kontinu di [0, \infty).

Determine the values of a and b to make the function f(x) continuous on the interval [0, ∞). This solution involves using limits and factoring techniques for a piecewise function. Suitable for students in Grade 11-12.

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