If `2f (x) + 3f (-x) = x^2 - x +1`, then the value of f'(1) is equal to

Options

A. `3/5`
B. `4/5`
C. `6/5`
D. `7/5`

Answer

To solve the equation \(2f(x) + 3f(-x) = x^2 - x + 1\) and find the value of \(f'(1)\), we will follow these steps:

### Step 1: Substitute \(x = 1\)
We start by substituting \(x = 1\) into the equation.

\[ 2f(1) + 3f(-1) = 1^2 - 1 + 1 \]
\[ 2f(1) + 3f(-1) = 1 \]

This gives us our first equation:
\[ (1) \quad 2f(1) + 3f(-1) = 1 \]

### Step 2: Substitute \(x = -1\)
Next, we substitute \(x = -1\) into the same equation.

\[ 2f(-1) + 3f(1) = (-1)^2 - (-1) + 1 \]
\[ 2f(-1) + 3f(1) = 1 + 1 + 1 = 3 \]

This gives us our second equation:
\[ (2) \quad 2f(-1) + 3f(1) = 3 \]

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \(2f(1) + 3f(-1) = 1\)
2. \(2f(-1) + 3f(1) = 3\)

We can solve these equations simultaneously. Let's multiply the first equation by 2 and the second equation by 3 to eliminate \(f(-1)\).

From equation (1):
\[ 4f(1) + 6f(-1) = 2 \]

From equation (2):
\[ 6f(-1) + 9f(1) = 9 \]

### Step 4: Subtract the equations
Now we subtract the first modified equation from the second modified equation:

\[ (6f(-1) + 9f(1)) - (4f(1) + 6f(-1)) = 9 - 2 \]
\[ (9f(1) - 4f(1)) = 7 \]
\[ 5f(1) = 7 \]
\[ f(1) = \frac{7}{5} \]

### Step 5: Find \(f(-1)\)
Now we can substitute \(f(1) = \frac{7}{5}\) back into one of the original equations to find \(f(-1)\). We can use equation (1):

\[ 2\left(\frac{7}{5}\right) + 3f(-1) = 1 \]
\[ \frac{14}{5} + 3f(-1) = 1 \]
\[ 3f(-1) = 1 - \frac{14}{5} \]
\[ 3f(-1) = \frac{5 - 14}{5} = \frac{-9}{5} \]
\[ f(-1) = \frac{-9}{15} = -\frac{3}{5} \]

### Step 6: Find \(f'(1)\)
To find \(f'(1)\), we need to differentiate the original equation with respect to \(x\):

\[ \frac{d}{dx}(2f(x) + 3f(-x)) = \frac{d}{dx}(x^2 - x + 1) \]
Using the chain rule:
\[ 2f'(x) - 3f'(-x) = 2x - 1 \]

Now substitute \(x = 1\):
\[ 2f'(1) - 3f'(-1) = 2(1) - 1 \]
\[ 2f'(1) - 3f'(-1) = 1 \]

We need another equation to solve for \(f'(1)\) and \(f'(-1)\). We can differentiate again using \(x = -1\):
\[ 2f'(-1) - 3f'(1) = 2(-1) - 1 \]
\[ 2f'(-1) - 3f'(1) = -2 - 1 = -3 \]

Now we have a new system of equations:
1. \(2f'(1) - 3f'(-1) = 1\)
2. \(2f'(-1) - 3f'(1) = -3\)

### Step 7: Solve for \(f'(1)\) and \(f'(-1)\)
From the first equation, we can express \(f'(-1)\):
\[ 3f'(-1) = 2f'(1) - 1 \implies f'(-1) = \frac{2f'(1) - 1}{3} \]

Substituting into the second equation:
\[ 2\left(\frac{2f'(1) - 1}{3}\right) - 3f'(1) = -3 \]
\[ \frac{4f'(1) - 2}{3} - 3f'(1) = -3 \]
Multiplying through by 3 to eliminate the fraction:
\[ 4f'(1) - 2 - 9f'(1) = -9 \]
\[ -5f'(1) = -7 \implies f'(1) = \frac{7}{5} \]

### Final Answer
Thus, the value of \(f'(1)\) is:
\[ \boxed{\frac{7}{5}} \]