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Consider the following functions:

r[x_] := r0 Exp[(k x)/(r1/(r1 + r[x]))];
f[x_] := (k x)/(r1/(r1 + r[x]));

and that

r1=1; r0=1; k=1;

I'd like to plot the function f[x]:

Plot[(k x)/(r1/(r1 + r[x])),{x,-1,1}]

How can the recursion relation can be dealt with?

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For recursion relations we can use RSolve (this case can be solved algebraically however, see comments):

sol1 = RSolve[r[x] == r0 Exp[(k x)/(r1/(r1 + r[x]))], r[x], x]

to plot:

Plot[(k x)/(r1/(r1 + r[x])) /. sol1 /. {r1 -> 1, r0 -> 1, 
   k -> 1}, {x, -1, 1}]

enter image description here

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    $\begingroup$ You will however have this warning: Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. $\endgroup$ – Ruud3.1415 Jan 22 '18 at 13:22
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    $\begingroup$ And in this specific case, there is no need for recursion so sol2 = Solve[r[x] == r0 Exp[(k x)/(r1/(r1 + r[x]))], r[x]] will give you the exact same thing algebraically $\endgroup$ – Ruud3.1415 Jan 22 '18 at 13:28

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