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The problem asks for the minimum value of the function \( f(x) = x^3 - 30x \) on the interval \( [2, 19] \). We will follow these steps:

1. **Find the derivative** \( f'(x) \) to identify critical points.
2. **Set the derivative equal to 0** to find potential critical points inside the interval.
3. **Evaluate the function** at the critical points and endpoints of the interval \( [2, 19] \).
4. **Determine the minimum value** among these points.

---

### Step 1: Derivative of \( f(x) = x^3 - 30x \)
Using the power rule:
\[
f'(x) = 3x^2 - 30
\]

### Step 2: Set the derivative to 0
\[
3x^2 - 30 = 0
\]
\[
3x^2 = 30 \quad \Rightarrow \quad x^2 = 10 \quad \Rightarrow \quad x = \pm \sqrt{10}
\]
Since we are only interested in the interval \( [2, 19] \), we take the positive solution:
\[
x = \sqrt{10} \approx 3.16
\]
This is a valid critical point within the interval.

---

### Step 3: Evaluate \( f(x) \) at Critical Points and Endpoints
- At \( x = 2 \):
  \[
  f(2) = 2^3 - 30(2) = 8 - 60 = -52
  \]

- At \( x = \sqrt{10} \):
  \[
  f(\sqrt{10}) = (\sqrt{10})^3 - 30\sqrt{10} = 10\sqrt{10} - 30\sqrt{10} = -20\sqrt{10}
  \]

- At \( x = 19 \):
  \[
  f(19) = 19^3 - 30(19) = 6859 - 570 = 6289
  \]

---

### Step 4: Identify the Minimum Value
We now compare the values:
- \( f(2) = -52 \)
- \( f(\sqrt{10}) = -20\sqrt{10} \approx -63.25 \)
- \( f(19) = 6289 \)

The smallest value is \( f(\sqrt{10}) = -20\sqrt{10} \).

---

### Final Answer
The minimum value of \( f(x) \) on the interval \( [2, 19] \) is:
\[
\boxed{-20\sqrt{10}}
\]
Thus, the correct option is **B**.

---

Do you have any questions or need further clarification?  
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### 5 Related Questions:
1. What does it mean when a critical point occurs outside the interval of interest?
2. How do we know whether a critical point is a minimum or maximum?
3. How would the solution change if the interval were \( [-5, 19] \)?
4. What is the geometric significance of a cubic function like this one?
5. How can we confirm that the minimum value found is indeed global on the interval?

### Tip:
Always include endpoints when finding the global minimum or maximum on a **closed interval**!

Giá trị nhỏ nhất của hàm số f(x) = x^3 - 30x trên đoạn [2;19] bằng?

Find the minimum value of the function f(x) = x^3 - 30x on the interval [2, 19]. This calculus problem involves finding critical points using derivatives and evaluating the function at critical points and endpoints to determine the global minimum. Suitable for high school students in grades 10-12.

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