A function y = f ( x) has a second order derivative f''(x) = 6 (x-1). If the graph passes through the point (2, 1) and at this point tangent to the graph is y = 3x - 1, then function is :

Options

A. `(x-1)^3`
B. `(x-1)^2`
C. `(x+1)^3`
D. `(x+1)^2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Integrate the second derivative
Given the second derivative \( f''(x) = 6(x - 1) \), we integrate it to find the first derivative \( f'(x) \).

\[ f'(x) = \int f''(x) \, dx = \int 6(x - 1) \, dx = 6 \left( \frac{x^2}{2} - x \right) + C_1 = 3x^2 - 6x + C_1 \]

### Step 2: Use the tangent condition
We know that at the point \( (2, 1) \), the tangent line is given by \( y = 3x - 1 \). The slope of this tangent line is 3, which means:

\[ f'(2) = 3 \]

Substituting \( x = 2 \) into the first derivative:

\[ f'(2) = 3(2^2) - 6(2) + C_1 = 3 \]
\[ 12 - 12 + C_1 = 3 \implies C_1 = 3 \]

Thus, the first derivative is:

\[ f'(x) = 3x^2 - 6x + 3 \]

### Step 3: Integrate the first derivative
Next, we integrate \( f'(x) \) to find \( f(x) \):

\[ f(x) = \int f'(x) \, dx = \int (3x^2 - 6x + 3) \, dx = x^3 - 3x^2 + 3x + C_2 \]

### Step 4: Use the point condition
We know that the function passes through the point \( (2, 1) \):

\[ f(2) = 1 \]

Substituting \( x = 2 \):

\[ f(2) = (2)^3 - 3(2)^2 + 3(2) + C_2 = 1 \]
\[ 8 - 12 + 6 + C_2 = 1 \implies 2 + C_2 = 1 \implies C_2 = -1 \]

Thus, the function is:

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

### Step 5: Final form
The final form of the function is:

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