If `f(x)=mx+c`,`f(0)=f'^`(0)=1` then `f(2)=`

Options

A. 1
B. 2
C. 3
D. -3

Explanation

To solve the problem step by step, we start with the function given:

### Step 1: Write down the function
The function is given as:
\[ f(x) = mx + c \]

### Step 2: Differentiate the function
To find the derivative of the function, we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = m \]

### Step 3: Use the condition \( f(0) = 1 \)
We know that \( f(0) = 1 \). Substituting \( x = 0 \) into the function:
\[ f(0) = m(0) + c = c \]
Since \( f(0) = 1 \), we have:
\[ c = 1 \]

### Step 4: Use the condition \( f'(0) = 1 \)
We also know that \( f'(0) = 1 \). Since \( f'(x) = m \), we can substitute \( x = 0 \):
\[ f'(0) = m = 1 \]

### Step 5: Substitute the values of \( m \) and \( c \) back into the function
Now that we have \( m = 1 \) and \( c = 1 \), we can write the function:
\[ f(x) = 1x + 1 = x + 1 \]

### Step 6: Find \( f(2) \)
To find \( f(2) \), we substitute \( x = 2 \) into the function:
\[ f(2) = 2 + 1 = 3 \]

### Final Answer
Thus, the value of \( f(2) \) is:
\[ \boxed{3} \]