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

Concept: By Rolle's Theorem, if a function is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ with $f(a) = f(b)$, there must exist at least one critical coordinate $c \in (a, b)$ such that $f'(c) = 0$. We can bound the first derivative away from this point by integrating the absolute limit bounds of the second derivative function using the Mean Value Theorem. Step 1: Apply Rolle's Theorem to locate a zero-slope critical point.
We are given that $f(x)$ is a twice differentiable function on the interval $[1, 3]$ such that $f(1) = f(3)$. According to Rolle's Theorem, because the boundary values are identical, there must exist at least one point $c \in (1, 3)$ where the first derivative drops to zero: \[ f'(c) = 0 \quad \text{for some } c \in (1, 3) \]

Step 2: Set up the absolute derivative bound using definite integration.
We are given the absolute constraint condition that the second derivative satisfies $|f''(x)| \le 2$ across the entire domain interval. Let us integrate this bound between our zero-slope reference coordinate $c$ and an arbitrary point $x \in [1, 3]$: \[ f'(x) - f'(c) = \int_{c}^{x} f''(t)\,dt \] Substitute $f'(c) = 0$ into the equation: \[ f'(x) = \int_{c}^{x} f''(t)\,dt \]

Step 3: Apply the absolute value integral inequality to maximize the range.
Taking the absolute value on both sides and applying the standard property that the magnitude of an integral is less than or equal to the integral of the magnitude ($|\int g| \le \int |g|$): \[ |f'(x)| = \left| \int_{c}^{x} f''(t)\,dt \right| \le \int_{\min(c,x)}^{\max(c,x)} |f''(t)|\,dt \] Substitute our maximum absolute upper bound constraint $|f''(t)| \le 2$ into the expression: \[ |f'(x)| \le 2 \cdot |x - c| \]

Step 4: Find the worst-case maximum distance to confirm the final constraint boundary.
Since $x$ and $c$ are both restricted to live inside the interval $[1, 3]$, the maximum possible absolute horizontal distance separating them occurs if one point is at an extreme endpoint and the other is at the opposite boundary. This spacing is strictly less than the total width of the domain: \[ |x - c| < 3 - 1 = 2 \quad \Rightarrow \quad |x - c| < 2 \] Substitute this distance threshold back into our derivative inequality: \[ |f'(x)| \le 2 \cdot |x - c| < 2 \cdot (2) = 4 \quad \Rightarrow \quad |f^{\prime}(x)| < 4 \] This matches option (D) perfectly.