Conditional expression

Question

How do I evaluate a conditional expression in order to convert the solutions that are in terms of $x$ to a new variable $c$ depending on $x$? The conventional method I have been using with eliminate doesn't work in this case, because the solution is not a simple inequality, but rather conditional.

 f[x_] := x^2 + c;
 g[x_] := f[f[x]];
 y = Solve[g'[x] == 1, x, Reals]

output is:

{{x -> ConditionalExpression[Root[-1 + 4 c #1 + 4 #1^3 &, 1], 
c > -(3/4) || c < -(3/4)]}, 
 {x ->  ConditionalExpression[Root[-1 + 4 c #1 + 4 #1^3 &, 2],  c < -(3/4)]}, 
 {x ->  ConditionalExpression[Root[-1 + 4 c #1 + 4 #1^3 &, 3],  c < -(3/4)]}}

 Eliminate[c == -x - x^2 && x == y[1], x]

Answer 1

I'm not sure if I understand your question correctly. But if you want to see the solutions depending on c this may be an approach:

f[x_] := x^2 + c;
g[x_] := f[f[x]];
y[c_] = Solve[g'[x] == 1, x, Reals]
xval[c_] := DeleteCases[y[c], {x -> Undefined}]

Test:

xval/@{-1,-3/4,1}
(*
{{{x -> Root[-1 - 4 #1 + 4 #1^3 &, 1]}, {x -> Root[-1 - 4 #1 + 4 #1^3 &, 2]}, 
  {x -> Root[-1 - 4 #1 + 4 #1^3 &, 3]}}, {}, {{x -> Root[-1 + 4 #1 + 4 #1^3 &, 1]}}}
*)

Answer 2

It's a bit unclear to me what you are asking, but if I interpret your question and your comment to Peter Breitfeld's answer correctly you want to solve the system

eqs = {g'[x] == 1, x^2 + x + c == 0}

for x and c. This can be done using Solve, e.g.

Solve[eqs, {x, c}]

(* output: {{x -> -(I/2), c -> 1/4 + I/2}, {x -> I/2, c -> 1/4 - I/2}} *)

Answer 3

How about the following?

c /. First@Solve[Expand[g'[x]] == 1, c]

Result is:

(1 - 4*x^3)/(4*x)