Let $f(x)$ be a twice differentiable function in $[1, 3]$ and $f(1) = f(3)$. Further if $|f''(x)| \leq 2$, then for all $x$ in $[1, 3]$

Since $f(1) = f(3)$, by Rolle's Theorem, there exists $c \in (1, 3)$ such that $f'(c) = 0$.

By Lagrange's Mean Value Theorem for $f'(x)$ on the interval between $x$ and $c$, there exists $d$ such that:

$f''(d) = \dfrac{f'(x) - f'(c)}{x - c}$

$\Rightarrow f'(x) = f''(d)(x - c)$

Taking absolute values on both sides:

$|f'(x)| = |f''(d)| |x - c|$

We are given that $|f''(x)| \leq 2$ for all $x \in [1, 3]$. Thus:

$|f'(x)| \leq 2 |x - c|$

Since $x \in [1, 3]$ and $c \in (1, 3)$, the maximum possible distance between $x$ and $c$ is strictly less than $3 - 1 = 2$.

$\Rightarrow |x - c| < 2$

Therefore, $|f'(x)| < 2 \times 2 = 4$.

Hence, $|f'(x)| < 4$ for all $x \in [1, 3]$.

Answer: $|f'(x)| < 4$