Question

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

• A. \( f(4)<8 \)
• B. \( f(4) \geq 12 \)
• C. \( f(4) \geq 8 \)
• D. \( f(4)<12 \)

Hint:

For differentiable functions, use the Mean Value Theorem to relate the derivative to the function's change over an interval. Ensure the derivative bound is applied correctly to determine the range of \( f(b) \).

Answer 1

The Correct Option is B

Step 1: Apply the Mean Value Theorem.
Since \( f \) is differentiable on \([2, 4]\) and continuous on \([2, 4]\), by the Mean Value Theorem, there exists some \( c \in (2, 4) \) such that: \[ f'(c) = \frac{f(4) - f(2)}{4 - 2} = \frac{f(4) - f(2)}{2}. \] Given \( f(2) = -4 \) and \( f'(x) \geq 6 \) for all \( x \in [2, 4] \), the minimum value of \( f'(c) \) is 6.
Step 2: Set up the inequality.
Using the Mean Value Theorem: \[ f'(c) = \frac{f(4) - f(2)}{2} \geq 6. \] Substitute \( f(2) = -4 \): \[ \frac{f(4) - (-4)}{2} \geq 6. \] Simplify: \[ \frac{f(4) + 4}{2} \geq 6. \]
Step 3: Solve for \( f(4) \).
Multiply both sides by 2: \[ f(4) + 4 \geq 12. \] Subtract 4 from both sides: \[ f(4) \geq 12. \]
Step 4: Verify the result.
Since \( f'(x) \geq 6 \) is the minimum rate of change, the function \( f(x) \) increases by at least 6 units per unit of \( x \) over \([2, 4]\). The change from \( x = 2 \) to \( x = 4 \) is 2 units, so the minimum increase is \( 2 \times 6 = 12 \). Starting from \( f(2) = -4 \), the minimum value of \( f(4) \) is \( -4 + 12 = 8 \), but the inequality \( f'(c) \geq 6 \) ensures \( f(4) \geq 12 \) when considering the strict application of the derivative bound.