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 $L(x)=f(a)+f^{\prime }(a)(x-a)$. In this case, $a=256$, $f(a)=4$, and $f^{\prime }(a)=1/256$.

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