I am having trouble solving this problem where I need to give conditions on $b$ and $c$ for the map $f \colon \mathbb{R} \rightarrow \mathbb{R}, f(x) = x^2 + bx + c$ to have a fixed point. Use these conditions to show that $f_c(x) = x^2 + c$ has a fixed point provided $c ≤ 1/4$.
What I did: Consider $f_c(x) = x^2 + c$ for $c\in \mathbb{R}$. We have 3 possible positions: when $c >1/4, c = 1/4, c <1/4$. $f_c(x)$ has no fixed points when $c>1/4$. When $x = 1/4$ it has unique fixed point at $x = 1/2$ (multiplicity of $2$ when you set function equal to $x$). When $c < 1/4$ it has a pair of fixed points where one point is attracting and the other is repelling.
$f \colon \mathbb{R} \rightarrow \mathbb{R}$ has a fixed point iff the function $f(x) = x$ has a solution in $\mathbb{R}$. So $f(x) = x^2 + bx + c$ becomes $x^2 + (b-1)x + c = 0$. How can I see if this function has a solution in $\mathbb{R}$? How do I use its conditions and relate it to function $f_c(x)$?
By the quadratic formula, $x^2 + (b-1)x + c = 0$ implies $$ x=\frac{-(b-1) \pm \sqrt{(b-1)^2-4c}}{2}. $$ To get a real solution, you need $(b-1)^2\geq 4c$. Regarding $f_c(x)=x^2+c$, it's the same thing with $b=0$, which results in $1\geq 4c$.