Find a linearization that will replace the function over an interval that includes the given point $x_{0}$. Center the linearization not at $x_{0}$ but at a nearby integer, $x=a$, at which the given function and its derivative are easy to evaluate.
\[
f(x)=\sqrt[4]{x}, x_{0}=255.7
\]
Set the center of the linearization as $x=$

Step 1 ：We are given the function \(f(x) = \sqrt[4]{x}\) and we are asked to find a linearization of this function at a point close to \(x_0 = 255.7\), but at an integer value where the function and its derivative are easy to evaluate.

Step 2 ：Looking at the function \(f(x) = \sqrt[4]{x}\), it's clear that it would be easiest to evaluate this function and its derivative at \(x=256\), because \(256\) is a perfect fourth power. So, we will set the center of the linearization as \(x=256\).

Step 3 ：The linearization of a function at a point \(x=a\) is given by the formula \(L(x) = f(a) + f'(a)(x-a)\). In this case, \(a=256\), \(f(a)=4\), and \(f'(a)=1/256\).

Step 4 ：Substituting these values into the formula, we get the linearization of the function at \(x=256\) as \(L(x) = \frac{x}{256} + 3\).

Step 5 ：\(\boxed{L(x) = \frac{x}{256} + 3}\) is the final answer.