You know that $\displaystyle R_2(x)=\frac{f^{\prime\prime\prime}(c)(x-2)^3}{3!}=\frac{2(x-2)^3}{6(c+3)^3}=\frac{(x-2)^3}{3(c+3)^3}$ with $1\le x\le 3$ and $c$ between 2 and $x$,
$\hspace{.6 in}$so $\displaystyle\big|R_2(x)\big|=\frac{|x-2|^3}{3(c+3)^3}\le\frac{1}{3(c+3)^3}<\frac{1}{3(4^3)}=\frac{1}{192}\;$
since $|x-2|\le1,\;$ and $\displaystyle1<c<3\implies 4<c+3<6\implies\frac{1}{4}>\frac{1}{c+3}\implies\frac{1}{(c+3)^3}<\frac{1}{4^3}$
The relationship between the trace of $A$ and the trace of $A^{-1}$ must depend on additional parameters other than $n$. Consider the example:
$$
A = \begin{bmatrix} 1 & 0 \\ 0 & \delta \end{bmatrix},
$$
where $\delta > 0$ is small. We observe that $\mathrm{tr}(A) = 1 + \delta$, $\mathrm{tr}(A^{-1}) = 1 + 1/\delta$, so $\mathrm{tr}(A)$ is nearly constant for small $\delta$ while $\mathrm{tr}(A^{-1})$ blows up.
The missing ingredient needed is the condition number. Generally speaking, the relationship between $A$ and $A^{-1}$ can be often characterized by the value of the condition number $\kappa$. For a general $n \times n$ matrix $A$, the condition number is defined by
$$
\kappa(A) = \frac{\sigma_{max}(A)}{\sigma_{min}(A)},
$$
where $\sigma_{max}$ and $\sigma_{min}$ are the largest and smallest singular values. If $A$ is invertible then this is equal to $\|A\| \|A^{-1}\|$. For an SPD matrix, this is equivalent to $\lambda_{max}(A)/\lambda_{min}(A)$.
Suppose $\kappa$ is large. Then, rescaling $A$ to have $\sigma_{max} = 1$, this means that $A$ has small norm but $A^{-1}$ has a large norm. This means that $A$ has a small trace, yet $A^{-1}$ has a large trace. Conversely, if $\kappa$ is close to $1$, then $A$ and $A^{-1}$ has similar norms, and their traces are also similar.
Going back to our original example, $\kappa(A) = 1/\delta$. Because the condition number of this matrix is large, then the trace of $A$ is close to $1$ while the trace of $A^{-1}$ is very large.
Now we can address your question. We have
\begin{align}
\mathrm{tr}(A) & = \lambda_1 + \lambda_2 + \ldots + \lambda_n \\
\mathrm{tr}(A^{-1}) & = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \ldots + \frac{1}{\lambda_n}.
\end{align}
Assuming $\lambda_1 \geq \ldots \geq \lambda_n > 0$, and utilizing the inequalities
$$
\frac{\lambda_1}{\kappa} \leq \lambda_j \le \lambda_n \kappa,
$$
we have
\begin{align}
\mathrm{tr}(A) \mathrm{tr}(A^{-1}) & \ge \left( \frac{ n \lambda_1}{\kappa} \right) \left( \frac{n}{\kappa \lambda_n} \right) = \frac{n^2}{\kappa}, \\
\mathrm{tr}(A) \mathrm{tr}(A^{-1}) & \le \left(n \lambda_n \kappa \right) \left( \frac{n \kappa}{ \lambda_1} \right) = n^2 \kappa.
\end{align}
In summary we have the inequalities
$$
\boxed{\frac{n^2}{\kappa} \le \mathrm{tr}(A) \mathrm{tr}(A^{-1}) \le n^2 \kappa}
$$
which can be converted into lower and upper bounds on the trace of $A$, in terms of the trace of $A^{-1}$, the condition number $\kappa$, and $n$.
Best Answer
No. The error is not bounded above by zero. The origin is not a local or global maximum for the error - it's the minimum instead. There's an absolute value in that definition, after all. [Now corrected]
Your polynomial is also incorrect - that's the wrong $x^4$ term. [Now corrected]
I suspect you're intended to use Taylor's theorem here. The theorem comes with estimates for the error of the series approximation. What does that error estimate from the theorem say in this case?