What does$ f(f(x))=0$ mean? I came across a question: If $f(x)=x^3-x+1$ then find the number of real distinct values of $f(f(x))=0$.
Here is what I interpreted the $f(f(x))$ as:
I assumed $a, b$ and $c$ to be the roots of $f(x)$, now if we put $a, b$ or $c$ in the $f(f(x))$ then it becomes $f(0)$ which will be equal to $1$.
I saw a solution where they differentiated the polynomial $f(x)$. They made the graph of $f(x)$ using first order derivative test. Then for $f(f(x))$, they put in $x=a, b, c$ (assumed roots of $f(x)$ ). Then we got three lines for $x=a$, a line below $-1$, for $b$ a line between $0$ and $1$ and for $c$, a line between $1$ and $3$. I am unable to understand why we put $a,b$ and $c$ as $x$ and then how did we get these ranges?


$f(f(x)) = (x^3 - x + 1)^3 - (x^3 - x + 1) + 1 = (x^3-x)^3+1+3(x^3-x)[x^3-x+1]-(x^3-x) = (x^3-x)^3+3(x^3-x)^2+3(x^3-x)-(x^3-x)+1 = (x^3-x)^3+3(x^3-x)^2+2(x^3-x) +1$
But given $f(f(x) = 0$ so $(x^3-x)^3+3(x^3-x)^2+2(x^3-x)+1 = 0$....eq.(1)
Put $x^3-x = y$ in eq.(1) we get $y^3+3y^2+2y+1 = 0$....eq.(2).
Firstly, solve eq.(2) then solve eq.(1).
