Let f(x) is a three degree polynomial for which `f ' (–1) = 0, f '' (1) = 0, f(–1) = 10, f(1) = 6` then local minima of f(x) exist at

Options

A. `x=3`
B. `x=2`
C. `x=1`
D. `x=-1`

Explanation

To solve the problem, we need to find the local minima of the cubic polynomial \( f(x) \) given the conditions:

1. \( f'(-1) = 0 \)
2. \( f''(1) = 0 \)
3. \( f(-1) = 10 \)
4. \( f(1) = 6 \)

### Step 1: Define the polynomial
Let \( f(x) = ax^3 + bx^2 + cx + d \).

### Step 2: Use the conditions to set up equations
From the conditions given, we can derive the following equations:

1. **From \( f(-1) = 10 \)**:
\[ f(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 10 \quad \text{(Equation 1)} \]

2. **From \( f(1) = 6 \)**:
\[ f(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = 6 \quad \text{(Equation 2)} \]

3. **From \( f'(-1) = 0 \)**:
First, we find the derivative:
\[ f'(x) = 3ax^2 + 2bx + c \]
Now substituting \( x = -1 \):
\[ f'(-1) = 3a(-1)^2 + 2b(-1) + c = 3a - 2b + c = 0 \quad \text{(Equation 3)} \]

4. **From \( f''(1) = 0 \)**:
We find the second derivative:
\[ f''(x) = 6ax + 2b \]
Now substituting \( x = 1 \):
\[ f''(1) = 6a(1) + 2b = 6a + 2b = 0 \quad \text{(Equation 4)} \]

### Step 3: Solve the system of equations
We have the following equations to solve:
1. \(-a + b - c + d = 10\) (Equation 1)
2. \(a + b + c + d = 6\) (Equation 2)
3. \(3a - 2b + c = 0\) (Equation 3)
4. \(6a + 2b = 0\) (Equation 4)

From Equation 4:
\[ 2b = -6a \implies b = -3a \]

Substituting \( b = -3a \) into Equations 1, 2, and 3:

**Substituting into Equation 1**:
\[ -a + (-3a) - c + d = 10 \implies -4a - c + d = 10 \quad \text{(Equation 5)} \]

**Substituting into Equation 2**:
\[ a + (-3a) + c + d = 6 \implies -2a + c + d = 6 \quad \text{(Equation 6)} \]

**Substituting into Equation 3**:
\[ 3a - 2(-3a) + c = 0 \implies 3a + 6a + c = 0 \implies 9a + c = 0 \implies c = -9a \quad \text{(Equation 7)} \]

### Step 4: Substitute \( c \) into Equations 5 and 6
Substituting \( c = -9a \) into Equations 5 and 6:

**From Equation 5**:
\[ -4a - (-9a) + d = 10 \implies 5a + d = 10 \implies d = 10 - 5a \quad \text{(Equation 8)} \]

**From Equation 6**:
\[ -2a + (-9a) + d = 6 \implies -11a + d = 6 \quad \text{(Equation 9)} \]

### Step 5: Solve for \( a \) and \( d \)
Substituting Equation 8 into Equation 9:
\[ -11a + (10 - 5a) = 6 \implies -16a + 10 = 6 \implies -16a = -4 \implies a = \frac{1}{4} \]

Now substituting \( a \) back to find \( b, c, d \):
\[ b = -3a = -3 \times \frac{1}{4} = -\frac{3}{4} \]
\[ c = -9a = -9 \times \frac{1}{4} = -\frac{9}{4} \]
\[ d = 10 - 5a = 10 - 5 \times \frac{1}{4} = 10 - \frac{5}{4} = \frac{40}{4} - \frac{5}{4} = \frac{35}{4} \]

### Step 6: Write the polynomial
Thus, the polynomial is:
\[ f(x) = \frac{1}{4}x^3 - \frac{3}{4}x^2 - \frac{9}{4}x + \frac{35}{4} \]

### Step 7: Find the critical points
To find the local minima, we set \( f'(x) = 0 \):
\[ f'(x) = 3 \cdot \frac{1}{4} x^2 + 2 \cdot -\frac{3}{4} x - \frac{9}{4} = \frac{3}{4} x^2 - \frac{3}{2} x - \frac{9}{4} \]
Setting this equal to zero:
\[ 3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \]
Factoring:
\[ (x - 3)(x + 1) = 0 \implies x = 3 \quad \text{or} \quad x = -1 \]

### Step 8: Determine the nature of critical points
We can use the second derivative test:
\[ f''(x) = 6ax + 2b = 6 \cdot \frac{1}{4} x + 2 \cdot -\frac{3}{4} = \frac{3}{2} x - \frac{3}{2} \]
Evaluating at \( x = 3 \):
\[ f''(3) = \frac{3}{2} \cdot 3 - \frac{3}{2} = \frac{9}{2} - \frac{3}{2} = \frac{6}{2} = 3 > 0 \quad \text{(local minima)} \]
Evaluating at \( x = -1 \):
\[ f''(-1) = \frac{3}{2} \cdot (-1) - \frac{3}{2} = -\frac{3}{2} - \frac{3}{2} = -3 < 0 \quad \text{(local maxima)} \]

### Conclusion
The local minima of \( f(x) \) exists at \( x = 3 \).