Rapid approximation of $\tanh(x)$

You are certainly aware that the hyperbolic tangent has an asymptote; since no polynomial ever had a horizontal asymptote, it stands to reason that a polynomial will always poorly approximate the qualitative behavior of the hyperbolic tangent.

One viable approach would be to consider the rational functions that come from truncations of the continued fraction for the hyperbolic tangent:

$$\tanh\,z=\cfrac{z}{1+\cfrac{z^2}{3+\cfrac{z^2}{5+\cdots}}}$$

Here for instance are plots comparing $\tanh\,z$ and the convergent

$$R(z)=\cfrac{z}{1+\cfrac{z^2}{3+\cfrac{z^2}{5+\cfrac{z^2}{7+\cfrac{z^2}{9+\cfrac{z^2}{11}}}}}}$$

The left plot shows that the rational approximant and the actual function are almost visually indistinguishable, while the right plot depicts the function $\tanh\,z-R(z)$.

One other possibility you can use in conjunction with rational function approximation is the use of argument reduction; in particular, the identity

$$\tanh\,z=\frac{2\tanh\frac{z}{2}}{1+\tanh^2\frac{z}{2}}$$

is very useful here. To sketch a possible algorithm: scale $z$ by $2^{-k}$, where $k$ is appropriately chosen such that $\dfrac{z}{2^k}$ is "small enough", evaluate the truncated continued fraction at this reduced argument, and then repeatedly apply the double argument identity $k$ times.

Again, in the interest of showing that there's always more than one way to skin a cat, I present a different set of rational functions that can be used instead of the continued fraction convergents I presented in my previous answer.

The idea is based on the Padé approximants of $\exp\,z$. More concretely, let the $(n,n)$ Padé approximant of $\exp\,z$ be represented by

$$\exp\,z\approx\frac{p_n(z)}{p_n(-z)}$$

where

$$p_n(z)=\sum_{j=0}^n\frac{\binom{n}{j}}{j!\binom{2n}{j}}z^j$$

From this, we find that we can approximate $\tanh\,z=\dfrac{\exp\,2z-1}{\exp\,2z+1}$ with a rational function like so:

$$\tanh\,z\approx\mathcal{T}_n(z)=\frac{p_n(z)^2-p_n(-z)^2}{p_n(z)^2+p_n(-z)^2}$$

For example,

$$\mathcal{T}_3(z)=\frac{z(10+z^2)(60+z^2)}{600+270z^2+11z^4+\frac{z^6}{24}}$$

Here are comparison plots for $\tanh\,z$ and $\mathcal{T}_3(z)$:

The left plot shows $\tanh\,z$ and $\mathcal{T}_3(z)$ together, while the right plot depicts the relative error function $1-\dfrac{\mathcal{T}_3(z)}{\tanh\,z}$. Note that the error is slightly smaller here than in the previous answer. One might still be able to improve on this approximant with a rational function with simple enough coefficients, but this is a start.

The best rational approximation to $\tanh(x)$ with numerator and denominator of degree 3 on the interval $[0, 3.1]$ (as provided by Maple's minimax function) is

(-.67436811832e-5+(.2468149110712040+(.583691066395175e-1+.3357335044280075e-1*x)*x)*x)/(.2464845986383725+(.609347197060491e-1+(.1086202599228572+.2874707922475963e-1*x)*x)*x)

This (call it $f(x)$) has maximum error .2735944241730870e-4, which is considerably less than 2^(-8).
On the interval $[-3.1, 3.1]$, use $\text{sgn}(x) f(|x|)$.