Let three degree polynomial function f(x) has local maximum at x = -1 and f(-1) = 2, f(3) = 18, f'(x) has a minima at x = 0, then :

Options

A. the distance between (-1,2) and (a,f(a)) where a denotes point where function has local max/min is `2sqrt5`
B. the function decreases from 1 to `2 sqrt5`
C. the function increases from 1 to `2sqrt5`
D. the function decreases from -1 to 1

Answer

To solve the problem, we need to find the polynomial function \( f(x) \) of degree 3 that satisfies the given conditions. Let's break it down step by step.

### Step 1: Assume the Form of the Polynomial
Since \( f(x) \) is a cubic polynomial, we can express it in the following form:
\[ f(x) = ax^3 + bx^2 + cx + d \]

### Step 2: Use Given Conditions
We have the following conditions:
1. \( f(-1) = 2 \)
2. \( f(3) = 18 \)
3. \( f'(-1) = 0 \) (local maximum at \( x = -1 \))
4. \( f'(0) \) has a minimum (which implies \( f''(0) > 0 \))

#### Condition 1: \( f(-1) = 2 \)
Substituting \( x = -1 \):
\[ f(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 2 \quad \text{(Equation 1)} \]

#### Condition 2: \( f(3) = 18 \)
Substituting \( x = 3 \):
\[ f(3) = a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d = 18 \quad \text{(Equation 2)} \]

### Step 3: Find the Derivative
The first derivative of \( f(x) \) is:
\[ f'(x) = 3ax^2 + 2bx + c \]
Since we have a local maximum at \( x = -1 \):
\[ f'(-1) = 3a(-1)^2 + 2b(-1) + c = 3a - 2b + c = 0 \quad \text{(Equation 3)} \]

### Step 4: Find the Second Derivative
The second derivative is:
\[ f''(x) = 6ax + 2b \]
Since \( f' \) has a minimum at \( x = 0 \), we have:
\[ f''(0) = 2b > 0 \quad \Rightarrow \quad b > 0 \]

### Step 5: Solve the Equations
Now we have three equations:
1. \( -a + b - c + d = 2 \) (Equation 1)
2. \( 27a + 9b + 3c + d = 18 \) (Equation 2)
3. \( 3a - 2b + c = 0 \) (Equation 3)

From Equation 3, we can express \( c \):
\[ c = 2b - 3a \quad \text{(Substituting into Equations 1 and 2)} \]

#### Substitute \( c \) into Equation 1:
\[ -a + b - (2b - 3a) + d = 2 \\ -a + b - 2b + 3a + d = 2 \\ 2a - b + d = 2 \quad \text{(Equation 4)} \]

#### Substitute \( c \) into Equation 2:
\[ 27a + 9b + 3(2b - 3a) + d = 18 \\ 27a + 9b + 6b - 9a + d = 18 \\ 18a + 15b + d = 18 \quad \text{(Equation 5)} \]

### Step 6: Solve Equations 4 and 5
Now we have:
1. \( 2a - b + d = 2 \) (Equation 4)
2. \( 18a + 15b + d = 18 \) (Equation 5)

Subtract Equation 4 from Equation 5:
\[ (18a + 15b + d) - (2a - b + d) = 18 - 2 \\ 16a + 16b = 16 \\ a + b = 1 \quad \text{(Equation 6)} \]

### Step 7: Substitute Back
From Equation 6, we can express \( b \):
\[ b = 1 - a \]

Substituting \( b \) into Equation 4:
\[ 2a - (1 - a) + d = 2 \\ 2a - 1 + a + d = 2 \\ 3a + d = 3 \quad \Rightarrow \quad d = 3 - 3a \quad \text{(Equation 7)} \]

### Step 8: Substitute \( b \) and \( d \) into \( c \)
Substituting \( b \) into \( c \):
\[ c = 2(1 - a) - 3a = 2 - 2a - 3a = 2 - 5a \]

### Step 9: Determine Values
Now we have:
- \( a \)
- \( b = 1 - a \)
- \( c = 2 - 5a \)
- \( d = 3 - 3a \)

We also know \( b > 0 \) implies \( 1 - a > 0 \) or \( a < 1 \).

### Step 10: Check Values
To find \( a \), we can use the condition \( f''(0) > 0 \):
\[ f''(0) = 2b = 2(1 - a) > 0 \quad \Rightarrow \quad 1 - a > 0 \quad \Rightarrow \quad a < 1 \]

### Step 11: Final Polynomial
Assuming \( a = 1 \):
- \( b = 0 \)
- \( c = -3 \)
- \( d = 0 \)

Thus, the polynomial is:
\[ f(x) = x^3 - 3x \]

### Step 12: Verify Conditions
1. \( f(-1) = 2 \) ✔
2. \( f(3) = 18 \) ✔
3. Local maximum at \( x = -1 \) ✔
4. Minimum of \( f' \) at \( x = 0 \) ✔

### Conclusion
The polynomial function that satisfies all the conditions is:
\[ f(x) = x^3 - 3x \]