If `f (x) +2 f (1-x) =x ^(2) +2 AA x in R and f (x)` is a differentiable function, then the value of `f'(8)` is

Answer

To solve the problem, we need to find the function \( f(x) \) from the given equation and then compute its derivative at \( x = 8 \). Let's go through the solution step by step.

### Step 1: Write down the given equation
We start with the equation:
\[ f(x) + 2f(1 - x) = x^2 + 2 \]

### Step 2: Substitute \( x \) with \( 1 - x \)
Next, we replace \( x \) with \( 1 - x \) in the original equation:
\[ f(1 - x) + 2f(x) = (1 - x)^2 + 2 \]
Calculating \( (1 - x)^2 \):
\[ (1 - x)^2 = 1 - 2x + x^2 \]
Thus, we have:
\[ f(1 - x) + 2f(x) = 1 - 2x + x^2 + 2 = x^2 - 2x + 3 \]

### Step 3: Write down the two equations
Now we have two equations:
1. \( f(x) + 2f(1 - x) = x^2 + 2 \) (Equation 1)
2. \( f(1 - x) + 2f(x) = x^2 - 2x + 3 \) (Equation 2)

### Step 4: Solve the system of equations
From Equation 1:
\[ f(x) + 2f(1 - x) = x^2 + 2 \]
From Equation 2:
\[ f(1 - x) + 2f(x) = x^2 - 2x + 3 \]

Now, we can express these equations in a more manageable form. Let's multiply Equation 1 by 2:
\[ 2f(x) + 4f(1 - x) = 2(x^2 + 2) \]
This gives us:
\[ 2f(x) + 4f(1 - x) = 2x^2 + 4 \]

Now we have:
1. \( 2f(x) + 4f(1 - x) = 2x^2 + 4 \) (Modified Equation 1)
2. \( f(1 - x) + 2f(x) = x^2 - 2x + 3 \) (Equation 2)

### Step 5: Subtract the equations
Now, we subtract Equation 2 from the modified Equation 1:
\[ (2f(x) + 4f(1 - x)) - (f(1 - x) + 2f(x)) = (2x^2 + 4) - (x^2 - 2x + 3) \]
This simplifies to:
\[ 3f(1 - x) = 2x^2 + 4 - x^2 + 2x - 3 \]
\[ 3f(1 - x) = x^2 + 2x + 1 \]
\[ f(1 - x) = \frac{x^2 + 2x + 1}{3} \]
\[ f(1 - x) = \frac{(x + 1)^2}{3} \]

### Step 6: Substitute back to find \( f(x) \)
Now we can substitute \( f(1 - x) \) back into Equation 1 to find \( f(x) \):
\[ f(x) + 2\left(\frac{(1 - x + 1)^2}{3}\right) = x^2 + 2 \]
\[ f(x) + 2\left(\frac{(2 - x)^2}{3}\right) = x^2 + 2 \]
Calculating \( (2 - x)^2 \):
\[ (2 - x)^2 = 4 - 4x + x^2 \]
Thus,
\[ f(x) + \frac{2(4 - 4x + x^2)}{3} = x^2 + 2 \]
\[ f(x) + \frac{8 - 8x + 2x^2}{3} = x^2 + 2 \]
Multiply everything by 3 to eliminate the fraction:
\[ 3f(x) + 8 - 8x + 2x^2 = 3x^2 + 6 \]
\[ 3f(x) = 3x^2 + 6 - 8 + 8x - 2x^2 \]
\[ 3f(x) = x^2 + 8x - 2 \]
\[ f(x) = \frac{x^2 + 8x - 2}{3} \]

### Step 7: Differentiate \( f(x) \)
Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\frac{x^2 + 8x - 2}{3}\right) = \frac{1}{3}(2x + 8) = \frac{2x + 8}{3} \]

### Step 8: Find \( f'(8) \)
Now we can find \( f'(8) \):
\[ f'(8) = \frac{2(8) + 8}{3} = \frac{16 + 8}{3} = \frac{24}{3} = 8 \]

### Final Answer
Thus, the value of \( f'(8) \) is:
\[ \boxed{8} \]