Answer :

Let's analyze the given function \( f(x) = 3x^3 + x^2 + 2x \).

To determine whether the function \( f(x) \) rises as \( x \) grows very small (i.e., as \( x \) approaches negative infinity), we need to investigate the behavior of the function as \( x \) approaches \( -\infty \).

### Step-by-Step Solution:

1. Analyze the leading term:
The term that will dominate the behavior of the polynomial for large positive or negative values of \( x \) is the term with the highest degree. In this case, it’s \( 3x^3 \).

2. Behavior as \( x \) approaches \( -\infty \):
- When \( x \) becomes very large in the negative direction (i.e., \( x \) approaches \( -\infty \)), the leading term \( 3x^3 \) will dominate the behavior of the function.
- Since the highest power is an odd degree (3), and the coefficient of \( x^3 \) is positive, \( 3x^3 \) will tend to \( -\infty \) as \( x \) approaches \( -\infty \).

3. Conclusion on the limit:
- Therefore, as \( x \) approaches \( -\infty \), the function \( f(x) \) also tends to \( -\infty \). This means that \( f(x) \) is not rising but rather decreasing without bound.

4. Identify the truth in the statement:
- The statement "The function \( f(x)=3 x^3+x^2+2 x \) rises as \( x \) grows very small" implies that the function increases as \( x \) approaches \( -\infty \).

Based on the analysis, the statement is indeed:

B. False

The function [tex]\( f(x) \)[/tex] does not rise as [tex]\( x \)[/tex] grows very small; instead, it decreases without bound.

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