If `f(x)=ax^(3)+bx^(2)+cx` have relative extrema x = 1 and at x = 5. If `int_(-1)^(1)f(x)dx=6` then :

Options

A. `a=-1`
B. `b=9`
C. `c=15`
D. `a=1`

Answer

To solve the given problem step by step, we will follow the outlined reasoning in the video transcript.

### Step 1: Find the derivative of \( f(x) \)
Given the function:
\[ f(x) = ax^3 + bx^2 + cx \]
We need to find the derivative \( f'(x) \) to locate the relative extrema:
\[ f'(x) = 3ax^2 + 2bx + c \]

### Step 2: Set the derivative equal to zero
Since we know that the relative extrema occur at \( x = 1 \) and \( x = 5 \), we can set up the following equations:
1. For \( x = 1 \):
\[ f'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c = 0 \quad \text{(Equation 1)} \]

2. For \( x = 5 \):
\[ f'(5) = 3a(5)^2 + 2b(5) + c = 75a + 10b + c = 0 \quad \text{(Equation 2)} \]

### Step 3: Set up the integral condition
We are given that:
\[ \int_{-1}^{1} f(x) \, dx = 6 \]
Calculating the integral:
\[ \int_{-1}^{1} (ax^3 + bx^2 + cx) \, dx = \left[ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 \right]_{-1}^{1} \]
Evaluating the limits:
\[ = \left( \frac{a}{4}(1)^4 + \frac{b}{3}(1)^3 + \frac{c}{2}(1)^2 \right) - \left( \frac{a}{4}(-1)^4 + \frac{b}{3}(-1)^3 + \frac{c}{2}(-1)^2 \right) \]
\[ = \left( \frac{a}{4} + \frac{b}{3} + \frac{c}{2} \right) - \left( \frac{a}{4} - \frac{b}{3} + \frac{c}{2} \right) \]
\[ = \frac{b}{3} + \frac{b}{3} = \frac{2b}{3} \]
Setting this equal to 6:
\[ \frac{2b}{3} = 6 \implies 2b = 18 \implies b = 9 \quad \text{(Equation 3)} \]

### Step 4: Substitute \( b \) into the equations
Now we substitute \( b = 9 \) into Equations 1 and 2.

1. From Equation 1:
\[ 3a + 2(9) + c = 0 \implies 3a + 18 + c = 0 \implies 3a + c = -18 \quad \text{(Equation 4)} \]

2. From Equation 2:
\[ 75a + 10(9) + c = 0 \implies 75a + 90 + c = 0 \implies 75a + c = -90 \quad \text{(Equation 5)} \]

### Step 5: Solve the system of equations
Now we have two equations (4 and 5):
1. \( 3a + c = -18 \)
2. \( 75a + c = -90 \)

Subtract Equation 4 from Equation 5:
\[ (75a + c) - (3a + c) = -90 + 18 \]
\[ 72a = -72 \implies a = -1 \]

### Step 6: Find \( c \)
Substituting \( a = -1 \) back into Equation 4:
\[ 3(-1) + c = -18 \implies -3 + c = -18 \implies c = -15 \]

### Summary of results
We have:
- \( a = -1 \)
- \( b = 9 \)
- \( c = -15 \)

### Final Answer
The values are:
- \( a = -1 \)
- \( b = 9 \)
- \( c = -15 \)