Problem Analysis:
- We are given a twice differentiable function $f(x)$ on the interval $[1, 3]$.
- Conditions provided: $f(1) = f(3)$ and $|f''(x)| \le 2$ for all $x \in [1, 3]$.
- The objective is to determine the constraint on the magnitude of the first derivative, $|f'(x)|$, for all $x$ within the interval $[1, 3]$.
Step 1: Apply Rolle's Theorem
The function $f(x)$ satisfies the conditions for Rolle's Theorem: it is continuous on $[1, 3]$, differentiable on $(1, 3)$, and the function values at the endpoints are equal ($f(1) = f(3)$). Therefore, Rolle's Theorem guarantees that there exists at least one point $c_1$ in the open interval $(1, 3)$ such that the first derivative is zero at that point, i.e., $f'(c_1) = 0$.
Step 2: Apply Mean Value Theorem to $f'(x)$
Since $f(x)$ is twice differentiable, its first derivative, $f'(x)$, is continuous and differentiable on $(1, 3)$. We can apply the Mean Value Theorem (MVT) to $f'(x)$ over an interval defined by $x$ and $c_1$. Let $x$ be any point in $[1, 3]$.
According to the MVT applied to $f'(t)$ over the interval between $x$ and $c_1$, there exists a point $c$ strictly between $x$ and $c_1$ such that: $f''(c) = \frac{f'(x) - f'(c_1)}{x - c_1}$
Substitute the result from Step 1 ($f'(c_1) = 0$) into the MVT equation: $f''(c) = \frac{f'(x) - 0}{x - c_1}$ $f''(c) = \frac{f'(x)}{x - c_1}$
Rearranging this equation to solve for $f'(x)$, we get: $f'(x) = f''(c) (x - c_1)$.
Step 3: Utilize the Bound on $f''(x)$
Now, consider the absolute value of $f'(x)$: $|f'(x)| = |f''(c)| |x - c_1|$
From the problem statement, we know that $|f''(x)| \le 2$ for all $x$ in $[1, 3]$. This implies $|f''(c)| \le 2$. Furthermore, $c$ lies strictly between $x$ and $c_1$. Since $x \in [1, 3]$ and $c_1 \in (1, 3)$, the distance $|x - c_1|$ is strictly less than the total length of the interval $[1, 3]$. Therefore, $|x - c_1| < (3 - 1) = 2$. Substitute these inequalities back into the expression for $|f'(x)|$:
$ |f'(x)| \le 2 \times |x - c_1| < 2 \times 2 $This calculation leads to the inequality:
$ |f'(x)| < 4 $This demonstrates that for any $x$ in the interval $[1, 3]$, the absolute value of the first derivative $f'(x)$ must be strictly less than 4.
Conclusion: Based on the derivation, the correct statement is $|f'(x)| < 4$, which corresponds to Option D.
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