What is the smallest value of m for which `f(x) = x^(2) + mx +5` is increasing in the interval `1 le x le 2`?

Options

A. m=0
B. `m=-1`
C. `m=-2`
D. `m=-3`

Answer

To find the smallest value of \( m \) for which the function \( f(x) = x^2 + mx + 5 \) is increasing in the interval \( 1 \leq x \leq 2 \), we need to analyze the derivative of the function.

### Step-by-Step Solution:

1. **Find the derivative of the function**:
The first step is to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^2 + mx + 5) = 2x + m \]

2. **Determine when the function is increasing**:
A function is increasing when its derivative is greater than or equal to zero. Therefore, we need:
\[ f'(x) \geq 0 \]
This leads to:
\[ 2x + m \geq 0 \]

3. **Evaluate the derivative at the endpoints of the interval**:
We need to check the values of \( f'(x) \) at the endpoints \( x = 1 \) and \( x = 2 \).

- For \( x = 1 \):
\[ f'(1) = 2(1) + m = 2 + m \geq 0 \implies m \geq -2 \]

- For \( x = 2 \):
\[ f'(2) = 2(2) + m = 4 + m \geq 0 \implies m \geq -4 \]

4. **Combine the inequalities**:
The function must satisfy both conditions:
\[ m \geq -2 \quad \text{and} \quad m \geq -4 \]
The more restrictive condition is \( m \geq -2 \).

5. **Conclusion**:
Therefore, the smallest value of \( m \) for which \( f(x) \) is increasing in the interval \( 1 \leq x \leq 2 \) is:
\[ \boxed{-2} \]