A function y = f(x) has a second-order derivative `f''(x) =6(x-1).` If its graph passed through the point (2,1) and at that point tangent to the graph is `y=3x-5,` then the value of f(0) is

Options

A. 1
B. `-1`
C. 2
D. 0

Answer

To solve the problem step by step, we need to find the value of \( f(0) \) given the second-order derivative \( f''(x) = 6(x - 1) \), the point through which the graph passes \( (2, 1) \), and the tangent line at that point \( y = 3x - 5 \).

### Step 1: Integrate the second derivative to find the first derivative

Given:
\[ f''(x) = 6(x - 1) \]

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

### Step 2: Use the tangent line to find \( C_1 \)

The tangent line at the point \( (2, 1) \) is given by \( y = 3x - 5 \). The slope of this line is \( 3 \), which means:
\[ f'(2) = 3 \]

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

Thus, we have:
\[ f'(x) = 3x^2 - 6x + 3 \]

### Step 3: Integrate the first derivative to find the function \( f(x) \)

Now, 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 \( (2, 1) \) to find \( C_2 \)

Since 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, we have:
\[ f(x) = x^3 - 3x^2 + 3x - 1 \]

### Step 5: Find \( f(0) \)

Now, substitute \( x = 0 \) to find \( f(0) \):
\[ f(0) = (0)^3 - 3(0)^2 + 3(0) - 1 = -1 \]

### Final Answer

The value of \( f(0) \) is:
\[ \boxed{-1} \]