Question

 For all x ∈ [0, 2024] assume that f (x) is differentiable. f (0) = −2 and f ′(x) ≥ 5. Then the least possible value of f (2024) is:

• A. \(10,120\)
• B. \(10,118\)
• C. \(10,122\)
• D. \(2024\)

Hint:

Apply the Mean Value Theorem and use the given inequality to find the least possible value.

Answer 1

The Correct Option is B

Step 1: Apply the Mean Value Theorem.
Since f(x) is differentiable on [0, 2024], by the Mean Value Theorem, there exists a c \(\in\) (0, 2024) such that:
f'(c) = \(\frac{f(2024) - f(0)}{2024 - 0}\)
f'(c) = \(\frac{f(2024) - (-2)}{2024}\)
f'(c) = \(\frac{f(2024) + 2}{2024}\)

Step 2: Use the given condition f'(x) \(\ge\) 5.
Since f'(x) \(\ge\) 5 for all x \(\in\) [0, 2024], we have f'(c) \(\ge\) 5.
\(\frac{f(2024) + 2}{2024} \ge 5\)

Step 3: Solve for f(2024).
f(2024) + 2 \(\ge\) 5 \(\times\) 2024
f(2024) + 2 \(\ge\) 10120
f(2024) \(\ge\) 10120 - 2
f(2024) \(\ge\) 10118

Step 4: Determine the least possible value of f(2024).
The least possible value of f(2024) is 10118.
Therefore, the least possible value of f(2024) is 10,118.