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 problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Finding the Derivative
Given the function:
\[ f(x) = ax^3 + bx^2 + cx \]

We first find the derivative:
\[ f'(x) = \frac{d}{dx}(ax^3) + \frac{d}{dx}(bx^2) + \frac{d}{dx}(cx) \]
Using the power rule:
\[ f'(x) = 3ax^2 + 2bx + c \]

### Step 2: Setting Derivative to Zero at Extrema
Since the function has relative extrema at \( x = 1 \) and \( x = 5 \), we set the derivative equal to zero at these points:
1. For \( x = 1 \):
\[ f'(1) = 3a(1^2) + 2b(1) + c = 0 \]
This simplifies to:
\[ 3a + 2b + c = 0 \] (Equation 1)

2. For \( x = 5 \):
\[ f'(5) = 3a(5^2) + 2b(5) + c = 0 \]
This simplifies to:
\[ 75a + 10b + c = 0 \] (Equation 2)

### Step 3: Subtracting the Equations
Now we subtract Equation 1 from Equation 2 to eliminate \( c \):
\[ (75a + 10b + c) - (3a + 2b + c) = 0 \]
This simplifies to:
\[ (75a - 3a) + (10b - 2b) = 0 \implies 72a + 8b = 0 \]
Dividing by 8:
\[ 9a + b = 0 \implies b = -9a \quad (Equation 3) \]

### Step 4: Using the Integral Condition
We are given the condition:
\[ \int_{-1}^{1} f(x) \, dx = 6 \]
Calculating the integral:
\[ \int_{-1}^{1} (ax^3 + bx^2 + cx) \, dx = \int_{-1}^{1} ax^3 \, dx + \int_{-1}^{1} bx^2 \, dx + \int_{-1}^{1} cx \, dx \]
The integrals evaluate as follows:
1. \(\int_{-1}^{1} ax^3 \, dx = 0\) (since \(x^3\) is an odd function)
2. \(\int_{-1}^{1} bx^2 \, dx = b \cdot \left[ \frac{x^3}{3} \right]_{-1}^{1} = b \cdot \left( \frac{1}{3} - \frac{-1}{3} \right) = \frac{2b}{3}\)
3. \(\int_{-1}^{1} cx \, dx = 0\) (since \(x\) is an odd function)

Thus, we have:
\[ \frac{2b}{3} = 6 \]
Multiplying both sides by 3:
\[ 2b = 18 \implies b = 9 \]

### Step 5: Finding \(a\) and \(c\)
Substituting \(b = 9\) into Equation 3:
\[ 9a + 9 = 0 \implies 9a = -9 \implies a = -1 \]

Now substituting \(a = -1\) and \(b = 9\) into Equation 1:
\[ 3(-1) + 2(9) + c = 0 \implies -3 + 18 + c = 0 \implies c = -15 \]

### Final Values
Thus, we have:
- \(a = -1\)
- \(b = 9\)
- \(c = -15\)

### Summary
The values of \(a\), \(b\), and \(c\) are:
- \(a = -1\)
- \(b = 9\)
- \(c = -15\)