Let `f(x)` be a cubic polynomial such that it has maxima at x=-1 , min at x=1 , `int_-1^1 f(x)dx=18 , f(2)=10` then find the sum of coff. of `f(x)`

Answer

To solve the problem, we need to find the cubic polynomial \( f(x) \) given certain conditions. Let's go through the steps systematically.

### Step 1: Understanding the Conditions
We know that:
- \( f(x) \) is a cubic polynomial.
- It has a maximum at \( x = -1 \) and a minimum at \( x = 1 \).
- The integral from \(-1\) to \(1\) of \( f(x) \) is \( 18 \).
- \( f(2) = 10 \).

### Step 2: Finding the Derivative
Since \( f(x) \) is a cubic polynomial, its derivative \( f'(x) \) will be a quadratic polynomial. Given the critical points (maximum and minimum), we can express \( f'(x) \) as:
\[ f'(x) = k(x + 1)(x - 1) = k(x^2 - 1) \]
where \( k \) is a constant.

### Step 3: Integrating the Derivative
To find \( f(x) \), we integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int k(x^2 - 1) \, dx = k\left(\frac{x^3}{3} - x\right) + C \]
Thus, we can write:
\[ f(x) = \frac{k}{3} x^3 - kx + C \]

### Step 4: Using the Given Conditions
1. **Condition from the Integral:**
We know:
\[ \int_{-1}^{1} f(x) \, dx = 18 \]
Since \( f(x) \) can be split into odd and even functions, the odd parts will cancel out over symmetric limits. Therefore, we only need to consider the constant term:
\[ \int_{-1}^{1} f(x) \, dx = \int_{-1}^{1} \left(-kx + C\right) \, dx = 2C \]
Setting this equal to \( 18 \):
\[ 2C = 18 \implies C = 9 \]

2. **Condition from \( f(2) = 10 \):**
Now substituting \( C \) into \( f(x) \):
\[ f(x) = \frac{k}{3} x^3 - kx + 9 \]
We can find \( k \) using \( f(2) = 10 \):
\[ f(2) = \frac{k}{3}(2^3) - k(2) + 9 = 10 \]
Simplifying this:
\[ \frac{8k}{3} - 2k + 9 = 10 \]
\[ \frac{8k}{3} - \frac{6k}{3} + 9 = 10 \]
\[ \frac{2k}{3} + 9 = 10 \implies \frac{2k}{3} = 1 \implies 2k = 3 \implies k = \frac{3}{2} \]

### Step 5: Writing the Final Form of \( f(x) \)
Substituting \( k \) and \( C \) back into \( f(x) \):
\[ f(x) = \frac{3}{6} x^3 - \frac{3}{2} x + 9 = \frac{1}{2} x^3 - \frac{3}{2} x + 9 \]

### Step 6: Finding the Sum of Coefficients
The coefficients of \( f(x) \) are:
- Coefficient of \( x^3 \): \( \frac{1}{2} \)
- Coefficient of \( x^1 \): \( -\frac{3}{2} \)
- Constant term: \( 9 \)

Now, summing these coefficients:
\[ \text{Sum of coefficients} = \frac{1}{2} - \frac{3}{2} + 9 = -1 + 9 = 8 \]

### Final Answer
The sum of the coefficients of \( f(x) \) is \( \boxed{8} \).