The curve `y =f (x)` satisfies `(d^(2) y)/(dx ^(2))=6x-4 and f (x)` has a local minimum vlaue 5 when `x=1.` Then `f^(prime)(0)` is equal to :

Options

A. 1
B. 0
C. 5
D. None of these

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the necessary equations.

### Step 1: Start with the given second derivative
We are given that the second derivative of the function \( f(x) \) is:
\[ \frac{d^2y}{dx^2} = 6x - 4 \]

### Step 2: Integrate to find the first derivative
To find the first derivative \( f'(x) \), we need to integrate the second derivative:
\[ f'(x) = \int (6x - 4) \, dx \]
Calculating the integral:
\[ f'(x) = 6 \cdot \frac{x^2}{2} - 4x + C_0 = 3x^2 - 4x + C_0 \]

### Step 3: Use the local minimum condition
We know that \( f(x) \) has a local minimum value of 5 when \( x = 1 \). This means:
\[ f'(1) = 0 \]
Substituting \( x = 1 \) into the first derivative:
\[ f'(1) = 3(1)^2 - 4(1) + C_0 = 0 \]
This simplifies to:
\[ 3 - 4 + C_0 = 0 \implies C_0 = 1 \]

### Step 4: Substitute \( C_0 \) back into the first derivative
Now we have:
\[ f'(x) = 3x^2 - 4x + 1 \]

### Step 5: Find \( f'(0) \)
To find \( f'(0) \):
\[ f'(0) = 3(0)^2 - 4(0) + 1 = 1 \]

### Final Answer
Thus, the value of \( f'(0) \) is:
\[ \boxed{1} \]