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.
Because \( f \) is differentiable on \([2, 4]\) and continuous on \([2, 4]\), the Mean Value Theorem guarantees a \( c \in (2, 4) \) such that: \[\nf'(c) = \frac{f(4) - f(2)}{4 - 2} = \frac{f(4) - f(2)}{2}.\n\] Given \( f(2) = -4 \) and \( f'(x) \geq 6 \) for all \( x \in [2, 4] \), the smallest value \( f'(c) \) can take is 6.
Step 2: Set up the inequality.
Using the Mean Value Theorem: \[\nf'(c) = \frac{f(4) - f(2)}{2} \geq 6.\n\] Substitute \( f(2) = -4 \): \[\n\frac{f(4) - (-4)}{2} \geq 6.\n\] Simplify: \[\n\frac{f(4) + 4}{2} \geq 6.\n\]
Step 3: Solve for \( f(4) \).
Multiply both sides by 2: \[\nf(4) + 4 \geq 12.\n\] Subtract 4 from both sides: \[\nf(4) \geq 12.\n\]
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 for every unit of \( x \) over \([2, 4]\). The change from \( x = 2 \) to \( x = 4 \) is 2 units, thus the minimum increase is \( 2 \times 6 = 12 \). Starting from \( f(2) = -4 \), the smallest possible value for \( f(4) \) is \( -4 + 12 = 8 \), but the inequality \( f'(c) \geq 6 \) implies \( f(4) \geq 12 \) when using the derivative bound.