If f(x) is a cubic polynomil which as local maximum at x=-1 . If f(2)=18,f(1)=-1 and f'(x) has minimum at x=0 then

Options

A. the distance between (-1,2) and (a,f(a)), where x=a is the point of local minima, is `2sqrt(5)`
B. f(x) is increasing for `x in [1,2sqrt(5)]`
C. f(x) has local minima at x=1
D. the value of f(0)=5

Answer

To solve the problem step by step, we will analyze the given information about the cubic polynomial \( f(x) \) and derive the necessary equations to find the required values.

### Step 1: Define the cubic polynomial
Assume the cubic polynomial is given by:
\[ f(x) = ax^3 + bx^2 + cx + d \]

### Step 2: Use the conditions provided
1. Since \( f(x) \) has a local maximum at \( x = -1 \), we know that:
\[ f'(-1) = 0 \]
2. We are given:
\[ f(2) = 18 \quad \text{and} \quad f(1) = -1 \]
3. The second derivative \( f''(x) \) has a minimum at \( x = 0 \):
\[ f''(0) = 0 \]

### Step 3: Find the first derivative
The first derivative of \( f(x) \) is:
\[ f'(x) = 3ax^2 + 2bx + c \]

### Step 4: Find the second derivative
The second derivative is:
\[ f''(x) = 6ax + 2b \]

### Step 5: Apply the condition for the second derivative
Setting \( f''(0) = 0 \):
\[ f''(0) = 2b = 0 \implies b = 0 \]

### Step 6: Update the polynomial
Now, substituting \( b = 0 \) into \( f(x) \):
\[ f(x) = ax^3 + cx + d \]

### Step 7: Apply the condition for the first derivative
Now, substituting \( b = 0 \) into the first derivative:
\[ f'(x) = 3ax^2 + c \]
Setting \( f'(-1) = 0 \):
\[ f'(-1) = 3a(-1)^2 + c = 3a + c = 0 \implies c = -3a \quad \text{(Equation 1)} \]

### Step 8: Use the values of \( f(1) \) and \( f(2) \)
1. For \( f(1) = -1 \):
\[ f(1) = a(1)^3 + c(1) + d = a - 3a + d = -2a + d = -1 \quad \text{(Equation 2)} \]
2. For \( f(2) = 18 \):
\[ f(2) = a(2)^3 + c(2) + d = 8a - 6a + d = 2a + d = 18 \quad \text{(Equation 3)} \]

### Step 9: Solve the system of equations
From Equation 2:
\[ d = -1 + 2a \quad \text{(Substituting into Equation 3)} \]
Substituting \( d \) into Equation 3:
\[ 2a + (-1 + 2a) = 18 \implies 4a - 1 = 18 \implies 4a = 19 \implies a = \frac{19}{4} \]

### Step 10: Find \( c \) and \( d \)
Using \( a \) to find \( c \):
\[ c = -3a = -3 \left(\frac{19}{4}\right) = -\frac{57}{4} \]
Now substituting \( a \) into Equation 2 to find \( d \):
\[ d = -1 + 2 \left(\frac{19}{4}\right) = -1 + \frac{38}{4} = \frac{34}{4} = \frac{17}{2} \]

### Step 11: Write the polynomial
The polynomial is:
\[ f(x) = \frac{19}{4}x^3 - \frac{57}{4}x + \frac{17}{2} \]

### Step 12: Determine local minima and maxima
1. The critical points are found by setting \( f'(x) = 0 \):
\[ 3ax^2 + c = 0 \implies 3 \left(\frac{19}{4}\right)x^2 - \frac{57}{4} = 0 \implies x^2 = \frac{57/4}{3(19/4)} = \frac{57}{57} = 1 \implies x = \pm 1 \]
- \( x = -1 \) is a local maximum.
- \( x = 1 \) is a local minimum.

### Step 13: Check the value at \( x = 0 \)
\[ f(0) = d = \frac{17}{2} \quad \text{(not equal to 5)} \]

### Step 14: Calculate the distance
The distance between points \( (-1, f(-1)) \) and \( (2, f(2)) \):
1. Calculate \( f(-1) \):
\[ f(-1) = \frac{19}{4}(-1)^3 - \frac{57}{4}(-1) + \frac{17}{2} = -\frac{19}{4} + \frac{57}{4} + \frac{34}{4} = \frac{72}{4} = 18 \]
2. The distance is:
\[ \text{Distance} = \sqrt{(2 - (-1))^2 + (f(2) - f(-1))^2} = \sqrt{(3)^2 + (18 - 18)^2} = \sqrt{9} = 3 \]

### Conclusion
The polynomial \( f(x) = \frac{19}{4}x^3 - \frac{57}{4}x + \frac{17}{2} \) has a local minimum at \( x = 1 \) and a local maximum at \( x = -1 \). The distance calculated is \( 3 \).