Tengo una imagen tikz y quería incluir una curva con la siguiente forma:
1/(x^2+1)
Puedo trazar la función para un dominio positivo, por ejemplo, [1,4]:
\draw[domain=1:4,smooth,variable=\x] plot ({\x},{1/(\x^2+1)});
Me gustaría poder trazar la función desde [-4,4].
Respuesta1
La solución es sustituir \x^2por\x*\x
Se vería así:
\begin{tikzpicture}
\fill[pattern=north west lines,opacity=.3] (-4,-1.11803*3) -- plot [domain=-4:4] ({\x},{-3/(\x*\x+2)-1.11803*3}) -- (4,-1.11803*3) -- cycle;
\draw[<-] (1.5,{-3/(1.5*1.5+2)-1.11803*3}) -- (2,-4.5) node[right] {$U(x)$};
\draw[<-] (-1.5,-3.6) -- (-2,-4.5) node[left] {$\displaystyle\int_{-\infty}^{\infty}\!\!\!\!\mbox{d}x \:U(x)$};
\draw[red,thick,domain=-4:4,smooth,samples=200,variable=\x] plot ({\x},{-3/(\x*\x+2)-1.11803*3});
\draw[red,dotted,thick,domain=-4:-5,smooth,samples=200,variable=\x] plot ({\x},{-3/(\x*\x+2)-1.11803*3});
\draw[red,dotted,thick,domain=4:5,smooth,samples=200,variable=\x] plot ({\x},{-3/(\x*\x+2)-1.11803*3});
\draw (0,0) circle (1.5cm);
\draw[->] (0,0) -- (0,-1.5) node[midway,left] {$R$};
\draw[->] (0,0) -- (1,-1.11803) node[midway,right] {$R$};
\draw[<->] (1,-1.11803) -- (1*3,-1.11803*3) node[midway,right] {$R$};
\draw[<->] (0,-1.5) -- (0,-1.11803*3) node[midway,right] {$z$};
\draw[-latex] (-5,-1.11803*3) -- (5,-1.11803*3);
\draw[|->|] (0,-1.11803*3-0.2) -- (3,-1.11803*3 -0.2) node[midway,below] {$x$};
\end{tikzpicture}
resultando como:
