lucas
Hi, is there a way to use enumerate like tabular so that the intervals in the snippet below are aligned with each other, like in the image? Thanks!


\usepackage[shortlabels,inline]{enumitem}

(...)

\begin{enumerate}[(i),before={\everymath{\displaystyle}}]
\item $$f(x)=x^{3}-x^{2}-8x+1$$;       $$[-2,2]$$
\item $$f(x)=x^{5}+x+1$$;              $$[-1,1]$$
\item $$f(x)=3x^{4}-8x^{3}+6x^{2}$$;    $$[-\frac12,\frac12]$$
\item $$f(x)=\frac{1}{x^{5}+x+1}$$;     $$[-\frac12,1]$$
\item $$f(x)=\frac{x+1}{x^{2}+1}$$;     $$[-1,\frac12]$$
\item $$f(x)=\frac{x}{x^{2}-1}$$;       $$[0,5]$$
\end{enumerate}


What I want:

![Captura de tela de 2021-02-11 15-20-08.png](/image?hash=61b056482a4e382ea4d9835d06a2dd3de5dce6a8e6da64024e9a736e3c12bc94)
samcarter
There are probably many possible approaches, here just a couple of them:

### Approach 1

one could make sure that the first terms in your enumerate all have the same width:


\usepackage[shortlabels,inline]{enumitem}
\begin{enumerate}[(i),before={\everymath{\displaystyle}}]
\item \makebox[4.5cm][l]{$$f(x)=x^{3}-x^{2}-8x+1$$;} on $$[-2,2]$$
\item \makebox[4.5cm][l]{$$f(x)=x^{5}+x+1$$;} on $$[-1,1]$$
\item \makebox[4.5cm][l]{$$f(x)=3x^{4}-8x^{3}+6x^{2}$$;} on $$[-\frac12,\frac12]$$
\item \makebox[4.5cm][l]{$$f(x)=\frac{1}{x^{5}+x+1}$$;} on $$[-\frac12,1]$$
\item \makebox[4.5cm][l]{$$f(x)=\frac{x+1}{x^{2}+1}$$;} on $$[-1,\frac12]$$
\item \makebox[4.5cm][l]{$$f(x)=\frac{x}{x^{2}-1}$$;} on $$[0,5]$$
\end{enumerate}


![Screen Shot 2021-02-11 at 22.24.55.png](/image?hash=3f9bb981d23b2a7ccafef2581c6479ff85b4c100a8175fed9a2cc42f9764100a)

### Approach 2

Use a table and emulate the enumerate label in the first column:


\newcounter{foo}
\newcommand{\myitem}{\stepcounter{foo}(\roman{foo})}

{
\everymath{\displaystyle}
\noindent\begin{tabular}{rll}
\setcounter{foo}{0}
\myitem & $$f(x)=x^{3}-x^{2}-8x+1$$ & on $$[-2,2]$$\\
\myitem & $$f(x)=x^{5}+x+1$$; & on $$[-1,1]$$\\
\myitem & $$f(x)=3x^{4}-8x^{3}+6x^{2}$$; & on $$[-\frac12,\frac12]$$\\
\myitem & $$f(x)=\frac{1}{x^{5}+x+1}$$; & on $$[-\frac12,1]$$\\
\myitem & $$f(x)=\frac{x+1}{x^{2}+1}$$; & on $$[-1,\frac12]$$\\
\myitem & $$f(x)=\frac{x}{x^{2}-1}$$; & on $$[0,5]$$\\
\end{tabular}
}


### Approach 3

Or you could use an align environment from the amsmath package:


\newcounter{foo}

\begin{flalign*}
\setcounter{foo}{0}
\myitem & f(x)=x^{3}-x^{2}-8x+1 && \text{on } [-2,2]\\
\myitem & f(x)=x^{5}+x+1; && \text{on }[-1,1]\\
\myitem & f(x)=3x^{4}-8x^{3}+6x^{2}; && \text{on } [-\frac12,\frac12]\\
\myitem & f(x)=\frac{1}{x^{5}+x+1}; && \text{on } [-\frac12,1]\\
\myitem & f(x)=\frac{x+1}{x^{2}+1}; && \text{on } [-1,\frac12]\\
\myitem & f(x)=\frac{x}{x^{2}-1}; && \text{on } [0,5]
\end{flalign*}


![Screen Shot 2021-02-11 at 22.36.44.png](/image?hash=6642f2824f96a3d38612e3fc7d047ae5029185b158d9b1a2fef8ecc0f7089c57)