Let `f(x)` be a cubic function such that `f'(1)=f''(2)=0`. If `x=1` is a point of local maxima of f(x), then the local minimum value of f(x) occurs at

Options

A. `x=0`
B. `x=2`
C. `x=4`
D. `x=3`

Answer

To solve the problem, we will follow these steps:

1. **Assume the Form of the Cubic Function**:
Let \( f(x) = ax^3 + bx^2 + cx + d \).

2. **Differentiate the Function**:
First, we find the first and second derivatives of \( f(x) \):
\[ f'(x) = 3ax^2 + 2bx + c \]
\[ f''(x) = 6ax + 2b \]

3. **Apply the Given Conditions**:
We know that \( f'(1) = 0 \) and \( f''(2) = 0 \).
Substituting \( x = 1 \) into \( f'(x) \):
\[ f'(1) = 3a(1^2) + 2b(1) + c = 3a + 2b + c = 0 \quad \text{(1)} \]
Substituting \( x = 2 \) into \( f''(x) \):
\[ f''(2) = 6a(2) + 2b = 12a + 2b = 0 \quad \text{(2)} \]

4. **Solve the System of Equations**:
From equation (2), we can express \( b \) in terms of \( a \):
\[ 2b = -12a \implies b = -6a \]
Now substitute \( b = -6a \) into equation (1):
\[ 3a + 2(-6a) + c = 0 \implies 3a - 12a + c = 0 \implies c = 9a \]

5. **Substitute Back into the Function**:
Now we have:
\[ b = -6a, \quad c = 9a \]
Thus, the function can be written as:
\[ f(x) = ax^3 - 6ax^2 + 9ax + d \]

6. **Find Critical Points**:
We need to find the critical points by setting \( f'(x) = 0 \):
\[ f'(x) = 3ax^2 - 12ax + 9a = 0 \]
Dividing by \( 3a \) (assuming \( a \neq 0 \)):
\[ x^2 - 4x + 3 = 0 \]
Factoring:
\[ (x - 3)(x - 1) = 0 \]
Thus, \( x = 1 \) and \( x = 3 \).

7. **Determine Local Maxima and Minima**:
Since \( x = 1 \) is given as a local maximum, the local minimum must occur at \( x = 3 \).

8. **Conclusion**:
Therefore, the local minimum value of \( f(x) \) occurs at:
\[ \boxed{3} \]