4. A fixed point of a function $f$ is a value of $x$ for which $f(x)=x$. Let $f$ be the quadratic function defined by $f(x)=x^{2}-c x+c$ where $c \in \mathbb{R}$. Find, in interval notation, the set consisting of all values of $c$ for which $f \circ f$ has four distinct fixed points.
Answer: $(-\infty,-1) \cup(3,+\infty)$
Solution: First, observe that both $x=c$ and $x=1$ are fixed points of $f$, and thus fixed points of $f \circ f$. Indeed, a fixed point of $f$ is a value of $x$ such that $f(x)-x=0$, and we have $f(x)-x=$ $x^{2}-(c+1) x+c=(x-1)(x-c)$. Moreover, $(f \circ f)(x)-x$ is a quartic polynomial in $x$, and its roots are the fixed points of $f \circ f$. Thus, it follows that $(f \circ f)(x)-x=(x-1)(x-c) g(x)$ for some quadratic polynomial $g$. Explicitly solving gives us $(f \circ f)(x)-x=(x-c)(x-1)\left(x^{2}-(c-1) x+1\right)$. Thus, we need only $x^{2}-(c-1) x+1$ to have two distinct roots, neither of which is equal to $c$ or 1 . This means that $(c-1)^{2}-4>0$, i.e. $(c+1)(c-3)>0$. Thus, we want $c3$. We check now that 1 and $c$ are not roots: we get if $x=1, x^{2}-(c-1) x+1=3-c \neq 0$ (since $\left.c \neq 3\right)$ and if $x=c, x^{2}-(c-1) x+1=c+1 \neq 0$ since $c \neq-1$.
Thus, all $c \in(-\infty,-1) \cup(3,+\infty)$ work.