Let `f(x)=x(x^(2)+mx+n)+2," for all" x neR and m, n in R. If ` Rolle's theorem holds for `f(x)at x=4//3 x in[1,2],` then `(m+n)"equal""_______".`

Answer

To solve the problem, we need to find the values of \( m \) and \( n \) such that the function \( f(x) = x(x^2 + mx + n) + 2 \) satisfies the conditions of Rolle's theorem on the interval [1, 2].

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = x(x^2 + mx + n) + 2 = x^3 + mx^2 + nx + 2 \]

2. **Check the conditions of Rolle's theorem**:
According to Rolle's theorem, \( f(1) = f(2) \).

3. **Calculate \( f(1) \)**:
\[ f(1) = 1^3 + m(1^2) + n(1) + 2 = 1 + m + n + 2 = m + n + 3 \]

4. **Calculate \( f(2) \)**:
\[ f(2) = 2^3 + m(2^2) + n(2) + 2 = 8 + 4m + 2n + 2 = 4m + 2n + 10 \]

5. **Set the equations equal**:
\[ f(1) = f(2) \implies m + n + 3 = 4m + 2n + 10 \]

6. **Rearranging the equation**:
\[ m + n + 3 - 4m - 2n - 10 = 0 \implies -3m - n - 7 = 0 \implies 3m + n + 7 = 0 \quad \text{(Equation 1)} \]

7. **Find the derivative \( f'(x) \)**:
\[ f'(x) = 3x^2 + 2mx + n \]

8. **Set the derivative to zero at \( x = \frac{4}{3} \)**:
\[ f'\left(\frac{4}{3}\right) = 3\left(\frac{4}{3}\right)^2 + 2m\left(\frac{4}{3}\right) + n = 0 \]
\[ 3 \cdot \frac{16}{9} + \frac{8m}{3} + n = 0 \implies \frac{48}{9} + \frac{8m}{3} + n = 0 \]
\[ \implies \frac{16 + 8m + 3n}{3} = 0 \implies 16 + 8m + 3n = 0 \quad \text{(Equation 2)} \]

9. **Solve the system of equations**:
We have two equations:
- \( 3m + n + 7 = 0 \) (Equation 1)
- \( 8m + 3n + 16 = 0 \) (Equation 2)

From Equation 1, we can express \( n \) in terms of \( m \):
\[ n = -3m - 7 \]

Substitute \( n \) into Equation 2:
\[ 8m + 3(-3m - 7) + 16 = 0 \]
\[ 8m - 9m - 21 + 16 = 0 \implies -m - 5 = 0 \implies m = -5 \]

Now substitute \( m \) back into Equation 1 to find \( n \):
\[ 3(-5) + n + 7 = 0 \implies -15 + n + 7 = 0 \implies n = 8 \]

10. **Calculate \( m + n \)**:
\[ m + n = -5 + 8 = 3 \]

### Final Answer:
\[ m + n = 3 \]