If both the critical points of `f(x) = ax^(3)+bx^(2) +cx +d` are -ve then

Options

A. `bc gt 0`
B. `bc lt 0`
C. `ab lt 0`
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) = ax^3 + bx^2 + cx + d \) and its critical points. The critical points are found by setting the first derivative equal to zero.

### Step-by-Step Solution:

1. **Find the First Derivative**:
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(ax^3 + bx^2 + cx + d) = 3ax^2 + 2bx + c \]

2. **Set the First Derivative to Zero**:
To find the critical points, we set the first derivative equal to zero:
\[ 3ax^2 + 2bx + c = 0 \]
This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \), where \( A = 3a \), \( B = 2b \), and \( C = c \).

3. **Identify the Roots**:
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-2b \pm \sqrt{(2b)^2 - 4(3a)(c)}}{2(3a)} \]
Simplifying this gives:
\[ x = \frac{-2b \pm \sqrt{4b^2 - 12ac}}{6a} \]

4. **Condition for Critical Points**:
We are given that both critical points are negative. This implies:
- The discriminant \( 4b^2 - 12ac \) must be non-negative for real roots.
- The sum of the roots (given by \( -\frac{B}{A} = -\frac{2b}{3a} \)) must be negative, which leads to:
\[ -\frac{2b}{3a} < 0 \implies \frac{b}{a} > 0 \]
This means \( b \) and \( a \) must have the same sign.

5. **Analyzing the Product of Roots**:
The product of the roots (given by \( \frac{C}{A} = \frac{c}{3a} \)) must also be positive for both roots to be negative:
\[ \frac{c}{3a} > 0 \implies c \text{ and } a \text{ must have the same sign.} \]

6. **Conclusion**:
From the above analysis, we can conclude:
- \( a \) and \( b \) must have the same sign.
- \( a \) and \( c \) must have the same sign.
- Therefore, \( a \), \( b \), and \( c \) must all be either positive or negative together.