Looking for peaks.

If $|f'(x)| \le 24$ , then $-24 \le f'(x) \le 24 \Rightarrow -24x + A \le f(x) \le 24x + B$ , where $A,B \in \mathbb{R}.$ Integrating the LHS linear equation first yields:

$\int_{0}^{1} -24x + A dx = -12x^2 + Ax|_{0}^{1} = 0 \Rightarrow A = 12$ , or $\boxed{-24x + 12}.$

Integrating the RHS linear equation prodcues:

$\int_{0}^{1} 24x + B dx = 12x^2 + Bx|_{0}^{1} = 0 \Rightarrow B = -12$ , or $\boxed{24x - 12}.$

Taking the first linear equation as $f(t)$ over $t \in [0,x]$ , we now take the absolute value of the integral over this interval to produce:

$|\int_{0}^{x} -24t + 12 dt| = |-12t^2 + 12t|_{0}^{x}| = |-12x^2 + 12x| = |-12(x-1/2)^2 + 3|$

which has zeros at $x = 0, 1$ and a maximum value of $k = 3$ . Thus, $8k^2 = 8 \cdot 3^2 = \boxed{72}.$

Likewise, taking the second linear equation as $f(t)$ over $t \in [0,x]$ , we now produce:

$|\int_{0}^{x} 24t - 12 dt| = |12t^2 - 12t|_{0}^{x}| = |12x^2 - 12x| = |12(x-1/2)^2 - 3|$

which also has zeros at $x = 0, 1$ and a maximum value of $k = 3$ . Thus, $8k^2 = 8 \cdot 3^2 = \boxed{72}.$