number of real roots of the equation $f(f(x)) = c$ The question is stated as follows: 
Let $f(x) = x^4 − 4x + 1$ for all real number $x$.
$\ $Determine, for each real number $c$, the
number of real roots of the equation $f(f(x)) = c.$

By calculating $f(f(x))$ explicitly it is easy to see that $f(f(x))=u^4+4u^3+6u^2-2$ with $u=x^4-4x$. The range of the outer function is $[-2, +\infty)$ so the number of real roots of the equation $f(f(x)) = c$ is zero if $c<-2$. How should I deal with case when $c \geq 2$?
 A: drew some graphs, fiddled with that. I don't see how you would know what to do without some sketches.
$$f(x) = x^4 - 4x + 1$$
$$g(x) = f(f(x)) = (x^4-4x+1)^4-(x^4-4x+1) + 1 \; , \; \; $$
then
$$   2+g(x) = x^2 \; (x^3-4)^2 \; \left( \; 2 +  (x^4 -4x+2)^2 \; \right) $$
where
$$ 2 +  (x^4 -4x+2)^2 \; > \; \; 0  $$
We see that $2+g(x) $  has double roots (local minima)  at $x=0$  and $x=  4^{1/3}$
You also need to know that there is a local max, the whole graph is a W  shape.
$$   -25+g(x) = (x-1)^2 \; (x^2 + 2x+3) \; \cdot \left( x^{12} - 12 x^9  + x^8 + 48 x^6 -8x^5  +3x^4  -64 x^3 + 16 x^2 -12x - 9 \right) $$
The factor of degree 12 has two real roots, as had to happen because of the W shape.  Indeed, taking
$$ q(x) = x^{12} - 12 x^9  + x^8 + 48 x^6 -8x^5  +3x^4  -64 x^3 + 16 x^2 -12x - 9 , $$
we find
$$ 9+q(x) = x (x^3-4) \left(x^8 - 8x^5 + x^4 + 16x^2 - 4x + 3   \right) , $$
where $$ x^8 - 8x^5 + x^4 + 16x^2 - 4x + 3  > 0$$
for some reason.
YES,
$$ x^8 - 8x^5 + x^4 + 16x^2 - 4x + 3  = \frac{11 + (2x^4 -8x+1)^2}{4}$$
I had it draw $y = \frac{g(x)}{10}$

Just to check,  the derivative
$$g'(x) = 16x^{15} - 208x^{12} + 48x^{11} + 960x^9 - 432x^8 + 48x^7 - 1792x^6 + 1152x^5 - 240x^4 + 1024x^3 - 768x^2 + 192x $$
factorws as
$$g'(x) = x (x-1) (x^3-4)  (x^2 + x+1) \left(x^8 - 8x^5 + 3x^4 + 16x^2 - 12x + 3 \right) $$
