我有一張 tikz 圖片，想要包含一條具有以下形式的曲線：
1/(x^2+1)
我可以繪製正域的函數，例如 [1,4]：
\draw[domain=1:4,smooth,variable=\x] plot ({\x},{1/(\x^2+1)});
我希望能夠繪製 [-4,4] 的函數。
答案1
解決方案是替換\x^2為\x*\x
它看起來像這樣：
\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}
結果為：
