# Plotting an implicit function using TikZ not displaying

I want to graph x^1/2-x and the implicit function (y^0.2-y)/2=x, but I'm having trouble with the second function. I'm not sure how to plot the function using tikz, and my attempt (below) does not seem to work.

> \usepackage{pgfplots}
>     \begin{document}
>     \tikzset{declare function={f(\x)=(\x)^(1/2)-\x;}}
>     \begin{tikzpicture}[scale=0.85]
>     \begin{axis}[domain=0:0.5,ytick={1},xtick={1,2,3,4},
>     xmin=0,ymin=0,ymax=0.5,xmax=0.5,mark=none,samples=100]
>           \draw[variable=\y,red]  plot ({(\y^0.2-\y)/2},{\y});
>     \end{axis}
>     \draw [black, dashed] plot coordinates {(-.1,2.85) (3.4,2.85)};
>     \draw [black, dashed] plot coordinates {(3.43,-.1) (3.43,2.8)};
>
>     \node at (3.43,-.3) {\small $1$};
>     \node at (6.5,-.3) {\small $b_1$};
>     \node at (-.23,2.85) {\small $1$};
>     \node at (-.23,5.4) {\small $b_2$};
>     \node at (5.8,3) {\small $b_2=r_2(b_1)$};
>     \node [text=red] at (4,5.4) {\small $b_1=r_1(b_2)$};
>     \end{document}

• This is because you pass all these arguments to \draw rather than \draw plot or \addplot. – user121799 Sep 12 '18 at 16:16
• Ah. I see. but are you sure my function will chart (y^0.2-y)/2=x? – pafnuti Sep 12 '18 at 16:19

Just pass these options to something appropriate like \addplot, or add plot.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\begin{document}
\tikzset{declare function={f(\x)=(\x)^(1/2)-\x;
g(\x)=(pow(\x,0.2)-\x)/2;}}
\begin{tikzpicture}[scale=0.85]
\begin{axis}[domain=0:0.5,ytick={1},xtick={1,2,3,4},
xmin=0,ymin=0,ymax=0.5,xmax=0.5,mark=none,samples=100]
\end{axis}
\draw [black, dashed] plot coordinates {(-.1,2.85) (3.4,2.85)};
\draw [black, dashed] plot coordinates {(3.43,-.1) (3.43,2.8)};
\begin{scope}[font=\small]
\node at (3.43,-.3) {$1$};
\node at (6.5,-.3) {$b_1$};
\node at (-.23,2.85) {$1$};
\node at (-.23,5.4) {$b_2$};
\node at (5.8,3) {$b_2=r_2(b_1)$};
\node [text=red] at (4,5.4) {$b_1=r_1(b_2)$};
\end{scope}
\end{tikzpicture}
\end{document}