Let f(x) be differentiable `AA x`. If f(1)=-2 and `f'(x) ge 2 AA x in x[1,6]`, then:

Options

A. `f[6] lt 5`
B. `f[6] =5`
C. `f[6] ge 8`
D. `f(6) lt 8`

Answer

To solve the problem, we need to find a condition for \( f(6) \) given that \( f(1) = -2 \) and \( f'(x) \geq 2 \) for \( x \) in the interval \([1, 6]\).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that \( f(1) = -2 \).
- The derivative \( f'(x) \) is greater than or equal to 2 for all \( x \) in the interval \([1, 6]\).

2. **Applying the Mean Value Theorem**:
- The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
- Here, let \( a = 1 \) and \( b = 6 \). Thus, we have:
\[ f'(c) = \frac{f(6) - f(1)}{6 - 1} \]

3. **Substituting Known Values**:
- Substituting \( f(1) = -2 \) into the equation gives:
\[ f'(c) = \frac{f(6) - (-2)}{5} = \frac{f(6) + 2}{5} \]

4. **Using the Condition on the Derivative**:
- Since we know that \( f'(x) \geq 2 \) for all \( x \) in \([1, 6]\), we can write:
\[ f'(c) \geq 2 \]
- Therefore, we have:
\[ \frac{f(6) + 2}{5} \geq 2 \]

5. **Solving the Inequality**:
- To eliminate the fraction, multiply both sides by 5:
\[ f(6) + 2 \geq 10 \]
- Subtracting 2 from both sides gives:
\[ f(6) \geq 8 \]

### Conclusion:
Thus, the condition we have found is:
\[ f(6) \geq 8 \]