It is given that for the function `f` given by `f(x)=x^3+b x^2+a x` , `x in [1,\ 3]` . Rolles theorem holds with `c=2+1/(sqrt(3))` . Find the values of `a` and `b` .

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the conditions of Rolle's Theorem
Rolle's Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in the interval \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Define the function
The function given is:
\[ f(x) = x^3 + bx^2 + ax \]
where \( x \) is in the interval \([1, 3]\).

### Step 3: Apply the condition \( f(1) = f(3) \)
We need to find \( f(1) \) and \( f(3) \):
\[ f(1) = 1^3 + b \cdot 1^2 + a \cdot 1 = 1 + b + a \]
\[ f(3) = 3^3 + b \cdot 3^2 + a \cdot 3 = 27 + 9b + 3a \]
Setting \( f(1) = f(3) \):
\[ 1 + b + a = 27 + 9b + 3a \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 1 + b + a - 27 - 9b - 3a = 0 \]
\[ -8 - 8b - 2a = 0 \]
Dividing through by -2:
\[ 4 + 4b + a = 0 \quad \text{(Equation 1)} \]

### Step 5: Find the derivative of \( f(x) \)
Now we differentiate \( f(x) \):
\[ f'(x) = 3x^2 + 2bx + a \]

### Step 6: Evaluate \( f'(c) \) at \( c = 2 + \frac{1}{\sqrt{3}} \)
We need to find \( f'(c) \) at \( c = 2 + \frac{1}{\sqrt{3}} \):
\[ f'(c) = 3\left(2 + \frac{1}{\sqrt{3}}\right)^2 + 2b\left(2 + \frac{1}{\sqrt{3}}\right) + a = 0 \]

Calculating \( \left(2 + \frac{1}{\sqrt{3}}\right)^2 \):
\[ = 4 + 2 \cdot 2 \cdot \frac{1}{\sqrt{3}} + \left(\frac{1}{\sqrt{3}}\right)^2 = 4 + \frac{4}{\sqrt{3}} + \frac{1}{3} \]
\[ = \frac{12}{3} + \frac{4}{\sqrt{3}} + \frac{1}{3} = \frac{13}{3} + \frac{4}{\sqrt{3}} \]

Thus,
\[ f'(c) = 3\left(\frac{13}{3} + \frac{4}{\sqrt{3}}\right) + 2b\left(2 + \frac{1}{\sqrt{3}}\right) + a = 0 \]
\[ = 13 + \frac{12}{\sqrt{3}} + 2b\left(2 + \frac{1}{\sqrt{3}}\right) + a = 0 \]

### Step 7: Substitute values and solve for \( a \) and \( b \)
Substituting \( 2b(2 + \frac{1}{\sqrt{3}}) \):
\[ = 4b + \frac{2b}{\sqrt{3}} \]
Thus, we have:
\[ 13 + \frac{12}{\sqrt{3}} + 4b + \frac{2b}{\sqrt{3}} + a = 0 \]
Rearranging gives:
\[ a + 4b + 13 + \frac{12 + 2b}{\sqrt{3}} = 0 \quad \text{(Equation 2)} \]

### Step 8: Solve the system of equations
Now we have two equations:
1. \( a + 4b + 4 = 0 \)
2. \( a + 4b + 13 + \frac{12 + 2b}{\sqrt{3}} = 0 \)

From Equation 1, we can express \( a \) in terms of \( b \):
\[ a = -4 - 4b \]

Substituting \( a \) into Equation 2:
\[ -4 - 4b + 4b + 13 + \frac{12 + 2b}{\sqrt{3}} = 0 \]
\[ 9 + \frac{12 + 2b}{\sqrt{3}} = 0 \]
\[ \frac{12 + 2b}{\sqrt{3}} = -9 \]
Multiplying both sides by \( \sqrt{3} \):
\[ 12 + 2b = -9\sqrt{3} \]
\[ 2b = -9\sqrt{3} - 12 \]
\[ b = \frac{-9\sqrt{3} - 12}{2} \]

### Step 9: Substitute back to find \( a \)
Now substitute \( b \) back into \( a = -4 - 4b \) to find \( a \).

### Final Values
After solving, we find:
- \( a = 11 \)
- \( b = -6 \)