To solve the problem, we need to determine the probability that the function \( f(x) = x^3 + ax^2 + bx + c \) is an increasing function, given that \( a, b, c \) are chosen by throwing a die three times.
### Step-by-Step Solution:
1. **Understanding the Function**:
The function \( f(x) \) is a cubic polynomial. For \( f(x) \) to be an increasing function, its derivative \( f'(x) \) must be non-negative for all \( x \).
2. **Finding the Derivative**:
We differentiate \( f(x) \):
\[
f'(x) = 3x^2 + 2ax + b
\]
For \( f(x) \) to be increasing, we need:
\[
f'(x) \geq 0 \quad \text{for all } x
\]
3. **Analyzing the Derivative**:
The quadratic \( 3x^2 + 2ax + b \) will be non-negative for all \( x \) if its discriminant is less than or equal to zero. The discriminant \( D \) of the quadratic \( Ax^2 + Bx + C \) is given by:
\[
D = B^2 - 4AC
\]
Here, \( A = 3 \), \( B = 2a \), and \( C = b \). Thus, the discriminant is:
\[
D = (2a)^2 - 4(3)(b) = 4a^2 - 12b
\]
For \( f'(x) \) to be non-negative for all \( x \):
\[
4a^2 - 12b \leq 0
\]
Simplifying this gives:
\[
a^2 \leq 3b
\]
4. **Choosing Values for a, b, and c**:
Since \( a, b, c \) are chosen by throwing a die, they can take values from 1 to 6. We need to find the pairs \( (a, b) \) that satisfy \( a^2 \leq 3b \).
5. **Finding Valid Combinations**:
We will check each possible value of \( a \) (from 1 to 6) and find the corresponding valid values of \( b \):
- If \( a = 1 \): \( 1^2 \leq 3b \) → \( 1 \leq 3b \) → \( b \geq 1 \) (valid for \( b = 1, 2, 3, 4, 5, 6 \)) → 6 options
- If \( a = 2 \): \( 2^2 \leq 3b \) → \( 4 \leq 3b \) → \( b \geq \frac{4}{3} \) → \( b = 2, 3, 4, 5, 6 \) → 5 options
- If \( a = 3 \): \( 3^2 \leq 3b \) → \( 9 \leq 3b \) → \( b \geq 3 \) → \( b = 3, 4, 5, 6 \) → 4 options
- If \( a = 4 \): \( 4^2 \leq 3b \) → \( 16 \leq 3b \) → \( b \geq \frac{16}{3} \) → \( b = 6 \) → 1 option
- If \( a = 5 \): \( 5^2 \leq 3b \) → \( 25 \leq 3b \) → No valid \( b \)
- If \( a = 6 \): \( 6^2 \leq 3b \) → \( 36 \leq 3b \) → No valid \( b \)
Now, we sum the valid combinations:
- For \( a = 1 \): 6 options
- For \( a = 2 \): 5 options
- For \( a = 3 \): 4 options
- For \( a = 4 \): 1 option
- For \( a = 5 \): 0 options
- For \( a = 6 \): 0 options
Total valid combinations for \( (a, b) \):
\[
6 + 5 + 4 + 1 = 16
\]
6. **Calculating Total Outcomes**:
Since \( a, b, c \) are chosen independently from a die, the total outcomes when throwing a die three times is:
\[
6 \times 6 \times 6 = 216
\]
7. **Calculating Probability**:
The probability that \( f(x) \) is an increasing function is given by:
\[
P(\text{increasing}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{16}{216} = \frac{2}{27}
\]
### Final Answer:
The probability that \( f(x) \) is an increasing function is \( \frac{2}{27} \).
