Question

Let \(f(x)\) be a twice differentiable function in \([1,3]\) and \(f(1)=f(3)\). Further if \(|f^{\prime\prime}(x)|\le2\), then for all \(x\) in \([1,3]\):

• A. $|f^{\prime}(x)|\ge4$
• B. $|f^{\prime}(x)|\le-1$
• C. $|f^{\prime}(x)|>2$
• D. $|f^{\prime}(x)|<4$

Hint:

Think of this dynamically: the second derivative acts as acceleration, and the first derivative acts as velocity. Since Rolle's theorem forces the velocity to hit 0 somewhere inside the interval, and the acceleration is capped at 2, the velocity can never pull away far enough to reach a speed of 4 within a time width of 2 units!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem applies the Mean Value Theorem (MVT). Since \( f(1) = f(3) \), by Rolle's Theorem, there exists at least one point \( c \in (1, 3) \) such that \( f'(c) = 0 \). The bound on the second derivative restricts how much the first derivative can change.
Step 2: Key Formula or Approach:
1. Rolle's Theorem: \( f'(c) = 0 \) for some \( c \in (1, 3) \).
2. Lagrange MVT on \( f'(x) \): \( \frac{f'(x) - f'(c)}{x - c} = f''(\xi) \).
Step 3: Detailed Explanation:
By Rolle's theorem, \( \exists c \in (1, 3) \) such that \( f'(c) = 0 \).
Now, for any \( x \in [1, 3] \), apply MVT to the function \( f'(t) \) on the interval between \( x \) and \( c \):
\[ |f'(x) - f'(c)| = |f''(\xi)| \cdot |x - c| \]
Since \( f'(c) = 0 \):
\[ |f'(x)| = |f''(\xi)| \cdot |x - c| \]
We are given \( |f''(x)| \le 2 \).
The maximum possible value of \( |x - c| \) in the interval \( [1, 3] \) occurs when \( x \) is at one boundary and \( c \) is at the other.
The length of the interval is \( 3 - 1 = 2 \).
So, \( |x - c|<2 \) (since \( c \) is strictly in the interior).
\[ |f'(x)| \le 2 \cdot |x - c|<2 \cdot 2 = 4 \]
Thus, \( |f'(x)|<4 \) for all \( x \in [1, 3] \).
Step 4: Final Answer:
The bound on the first derivative is \( |f'(x)|<4 \), which is option (D).