Find the root of this problem

$f(f(x))=0$

$∴ f(x)= f^-(0)$

$∴ x^3-3x+1 = f^-(0)$

$x^3-3x+1 - f^-(0)$ $=0$

$g(x)$ = $x^3-3x+1 - f^-(0)$ = 0

As $f(x)$ is decreasing in $(-1,1)$ and $f(1)$ $< 0$ , so graph of $f(x)$ will cut X-axis in three points $'a,b,c'$ ,

where $a<-1, 0<b<1, c>1$ . So $f^-(0) ='a,b,c'$ . Now, $g(x)$ = $x^3-3x+1 - f^-(0)$ $= 0$

if $f^-(0) = a ⇒ x^3-3x+1 - a= 0$ , so $g(1) = -2+1-a=-1-a$ . As, $a<-1 ⇒ -1-a >0$ , So $g(x)$ have only one root on negative X-axis , as it is decreasing in $(-1,1)$ and then increasing. Similarly for $f^-(0) = b ⇒ x^3-3x+1-b = 0$ , $g(1) = -1-b$ , here $0<b<1$ so $g(1) < 0$ and so it will have three roots and same for $f^-(0) = c$ Thus it will have total $3+3+1=7$ roots

The main idea over here is to find the critical points of the function $f(f(x))$ and then checking the sign of $f(f(x))$ at this critical points so as to get a rough picture of the plot of this graph.

Let us call $f(f(x))=g(x)$

Therefore $g(x) = f^3(x)-3f(x)+1$

$\Rightarrow g'(x)=3f'(x) \left(f^2(x)-1\right )$

Therefore for $g'(x)=0$ we have

• $f'(x)=0 \Rightarrow x \in \{-1,1\}$

• $f^2(x)=1 \Rightarrow x \in \{-2,-\sqrt3, 0, 1, \sqrt3\}$

Therefore the solution for the critical points of $g(x)$ is $x\in \{-2, -\sqrt3, -1, 0, 1, \sqrt3\}$

Finally

• $g(-2)> 0$

• $g(-\sqrt3) < 0$

• $g(-1)> 0$

• $g(0)< 0$

• $g(1)> 0$

• $g(\sqrt3)< 0$

Using the above data if we roughly try to plot $g(x)$ it is found to cut the $x-axis$ at $7$ distinct points meaning it has $7$ distinct real roots