Suppose `f(x)=x^(3)+ax^2+bx+c.a,b,c` are chosen respectively by throwing a die three times. Find the probability that f(x) is an increasing function.

Answer

To solve the problem of finding the probability that the function \( f(x) = x^3 + ax^2 + bx + c \) is an increasing function, we will follow these steps:

### Step 1: Determine the condition for \( f(x) \) to be increasing
The function \( f(x) \) is increasing if its derivative \( f'(x) \) is non-negative for all \( x \). The derivative is given by:

\[ f'(x) = 3x^2 + 2ax + b \]

For \( f'(x) \) to be non-negative for all \( x \), the quadratic \( 3x^2 + 2ax + b \) must either have no real roots or be a perfect square. This condition is satisfied when the discriminant \( D \) of the quadratic is less than or equal to zero:

\[ D = (2a)^2 - 4 \cdot 3 \cdot b \leq 0 \]

### Step 2: Simplify the discriminant condition
Calculating the discriminant:

\[ D = 4a^2 - 12b \leq 0 \]

This simplifies to:

\[ 4a^2 \leq 12b \quad \Rightarrow \quad a^2 \leq 3b \]

### Step 3: Determine the possible values of \( a \) and \( b \)
Since \( a \), \( b \), and \( c \) are chosen by throwing a die, their possible values are \( 1, 2, 3, 4, 5, 6 \).

### Step 4: Count the favorable outcomes
We need to count the pairs \( (a, b) \) such that \( a^2 \leq 3b \).

1. **If \( a = 1 \)**:
- \( 1^2 \leq 3b \) implies \( 1 \leq 3b \) or \( b \geq \frac{1}{3} \). Thus, \( b \) can be \( 1, 2, 3, 4, 5, 6 \) (6 options).

2. **If \( a = 2 \)**:
- \( 2^2 \leq 3b \) implies \( 4 \leq 3b \) or \( b \geq \frac{4}{3} \). Thus, \( b \) can be \( 2, 3, 4, 5, 6 \) (5 options).

3. **If \( a = 3 \)**:
- \( 3^2 \leq 3b \) implies \( 9 \leq 3b \) or \( b \geq 3 \). Thus, \( b \) can be \( 3, 4, 5, 6 \) (4 options).

4. **If \( a = 4 \)**:
- \( 4^2 \leq 3b \) implies \( 16 \leq 3b \) or \( b \geq \frac{16}{3} \). Thus, \( b \) can only be \( 6 \) (1 option).

5. **If \( a = 5 \)**:
- \( 5^2 \leq 3b \) implies \( 25 \leq 3b \) or \( b \geq \frac{25}{3} \). No valid \( b \) exists (0 options).

6. **If \( a = 6 \)**:
- \( 6^2 \leq 3b \) implies \( 36 \leq 3b \) or \( b \geq 12 \). No valid \( b \) exists (0 options).

### Step 5: Total favorable outcomes
Now, we sum the number of valid \( b \) values for each \( a \):

- 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 favorable pairs \( (a, b) \) = \( 6 + 5 + 4 + 1 + 0 + 0 = 16 \).

### Step 6: Calculate the total outcomes
The total number of outcomes when throwing a die three times (for \( a, b, c \)) is:

\[ 6 \times 6 \times 6 = 216 \]

### Step 7: Calculate the probability
The probability \( P \) that \( f(x) \) is an increasing function is given by:

\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{16}{216} = \frac{2}{27} \]

### Final Answer
The probability that \( f(x) \) is an increasing function is:

\[ \frac{2}{27} \]

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