Let f be a function which is continuous and differentiable for all real x. If `f(2)=-4 and f'(x) >= 6` for all `x in [2,4],` then

Options

A. `f(4)lt8`
B. `f(4)ge8`
C. `f(4)ge2`
D. none of these

Explanation

To solve the problem step by step, we will use the Mean Value Theorem (MVT) and the given conditions about the function \( f \).

### Step 1: Understand the Given Information
We know that:
- \( f \) is continuous and differentiable for all real \( x \).
- \( f(2) = -4 \).
- \( f'(x) \geq 6 \) for all \( x \) in the interval \([2, 4]\).

### Step 2: Apply the Mean Value Theorem
According to the Mean Value Theorem, there exists at least one \( c \) in the interval \((2, 4)\) such that:

\[ f'(c) = \frac{f(4) - f(2)}{4 - 2} \]

### Step 3: Substitute the Known Values
Substituting \( f(2) = -4 \) into the equation gives us:

\[ f'(c) = \frac{f(4) - (-4)}{2} = \frac{f(4) + 4}{2} \]

### Step 4: Use the Condition on the Derivative
Since we know that \( f'(x) \geq 6 \) for all \( x \) in \([2, 4]\), we can say:

\[ f'(c) \geq 6 \]

### Step 5: Set Up the Inequality
From the previous steps, we have:

\[ \frac{f(4) + 4}{2} \geq 6 \]

### Step 6: Solve the Inequality
Multiply both sides by 2 to eliminate the fraction:

\[ f(4) + 4 \geq 12 \]

Now, subtract 4 from both sides:

\[ f(4) \geq 8 \]

### Conclusion
Thus, we conclude that:

\[ f(4) \geq 8 \]

### Final Answer
The range of \( f(4) \) is \( f(4) \geq 8 \).

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