finding the continuous function $x $ on [-1, 1] find the continuous function $x $ on $[-1, 1]$ such that 
\begin{equation} 
\frac{|\int_{-1}^{0} x   \mathrm{d}t-\int_{0}^{1}x \mathrm{d}t|}{max_{[-1, 1]}|x|}\geq 2
\end{equation}

Such a continuous function does not exist!
If $M:=\max_{[-1, 1]}|x|>0$ and the equality holds then $x$ has to be constant in $(0,1]$, and $[-1,0)$, with different constants (because otherwise $\left|\int_{-1}^{0} x  dt-\int_{0}^{1}x dt\right|=0$).
Hence $x$ can not be continuous in $[-1,1]$. 
In fact, since $x$ is continuous in $[0,1]$ and in $[-1,0]$, if $x$ is not constant on $(0,1]$ and $[-1,0)$, then at least one of the integrals $\int_{-1}^{0} |x|  dt$ and $\int_{0}^{1}|x| dt$ is strictly less than $M$ and
$$2M=\left|\int_{-1}^{0} x  dt-\int_{0}^{1}x dt\right|\leq \int_{-1}^{0} |x|  dt+\int_{0}^{1}|x| dt< 2M.$$
Contradiction!
