Real Roots of a Triple Function

$\begin{array}{c}\ f(f (f (x) )) & = \left( (x^2-1)^2-1 \right) ^2-1 & \\ &= \left[(x^2-1)^2-1 + 1\right] \left[ (x^2-1)^2-1-1\right] & \\ & = \left(x-1\right)^2\left(x+1\right)^2 \left(x^2 -1-\sqrt{2}\right)\left(x^2-1+\sqrt{2}\right) & \end{array}$

The 2 linear terms contribute the real roots $1$ and $-1$ , the first quadratic contributes the real roots $-\sqrt{1+\sqrt{2}}$ and $\sqrt{1+\sqrt{2}}$ , and the last quadratic has 2 imaginary roots. Hence there are $4$ distinct real roots in all.

We want to solve $[(x^2-1)^2-1]^2 = 1$ .

Case (A): $(x^2-1)^2-1 = 1$ , so $(x^2-1)^2=2$ . Subcases

Case (A1): $x^2-1 = \sqrt{2}$ . $x^2 = 1 + \sqrt{2}$ . Since RHS is positive, we have 2 real solutions.

Case (A2): $x^2-1 = -\sqrt{2}$ . $x^2 = 1 - \sqrt{2}$ . Since RHS is negative, we have no real solutions.

Case (B): $(x^2-1)^2-1=-1$ . $x^2-1 = 0$ , $x^2 = 1$ . So two solutions $x=+1$ and $x=-1$

f(x)=xsquare -1 f(f(x))=(xquare-1)square-1=a f(f(f(x))=a square -1 given a suuare-1 =0 implies a square=1 implies a = + or -1 calculating for a=1 we get 2 roots which are x=+/- square root of (square root 2 + 1) and solving for a= -1 we get 2 more roots which are +/- 1.

f(x)=x^2-1 then f(f(x)=(x-1)^2-1 then f(f(fx)))=((x^2-1)^2-1)^2-1 we solve f(f(f(x)))=0 that is ((x^2-1)^2-1)^2-1=0 \Rightarrow ((x^2-1)^2-1)^2=1 \Rightarrow (x^2-1)^2-1=1 or (x^2-1)^2-1= -1 when (x^2-1)^2-1=1 \Rightarrow (x^2-1)^2 =2 \Rightarrow x^2-1 = \sqrt{2} or (x^2-1)^2 =-\sqrt{2} (invalid) we have x^2-1 = \sqrt{2} then x^2 = 1+\sqrt{2} \Rightarrow x= \pm \sqrt{1+\sqrt{2}} then we have two solutions when (x^2-1)^2-1= -1 \Rightarrow (x^2-1)^2 = 0 \Rightarrow x^2-1=0 \Rightarrow x^2=1 then x= \pm 1 we have two solutions so we have 4 distinct real roots.

$f(f(f(x)))$ $=f(f(x^2-1))$ $=f((x^2-1)^2-1)$ $=f(x^4-2x^2)$ $=(x^4-2x^2)^2-1$ $=((x^4-2x^2)+1)((x^4-2x^2)-1)$ $=(x^2-1)^2(x^4-2x^2-1)$ $=(x+1)^2(x-1)^2(x^4-2x^2-1)$ Since $f(f(f(x)))=0$ , then $x+1=0$ , $x+1=0$ , and $x^4-2x^2-1=0$ . Solving for $x$ , $x=-1, +1, \sqrt{1+\sqrt{2}}, -\sqrt{1+\sqrt{2}}, \sqrt{1-\sqrt{2}}, -\sqrt{1-\sqrt{2}}$ However, the last two values of x are imaginary. Hence, there are only four distinct real roots.

f(f(f(x)))=0 implies f(f(x))=1 or -1 Consider each case in turn f(f(x))=1 implies f(x)= $\pm$ $\sqrt(2)$

f(x)= $-\sqrt(2)$ has no solutions f(x)= $\sqrt(2)$ has 2 solutions

Similarly f(f(x))=-1 has only 1 solution i.e. f(x)=0 This has 2 solutions for x, making a total of 4

Lets write f(f(x))=g(x). Then f(g(x))=0 is same as $g^2(x)-1 = 0$ . This gives g(x)=1 or g(x)=-1 . Now, for first case, $f(f(x)) = f^2(x)-1=1$ . For second, $f^2(x)-1=-1$ . For first case $f(x)=+\sqrt{2}$ or $f(x)=-\sqrt{2}$ . Only one of these here is possible, $f(x)=+\sqrt{2}$ , since $1 - \sqrt{2}$ is less than 0. For second case, f(x)=0. They both give 2 solutions each.

fff(x)=ff(x^2 -1)

0=f((x^2 -1)^2 -1)

0=f(x^4 -2x^2 +1-1)

0=(x^4 -2x^2)^2 -1

0=x^8 - 4x^6 + 4x^4 -1

1=x^4 (x^4 - 4x^2 +4)

x=1 or

x^4 - 4x^2 + 4 = 1

x^4 - 4x^2 + 3 = 0

( x^2 -1 ) ( x^2 -3 ) = 0

x^2 = 1 or x^2 = 3

x = 1, -1 or x = \sqrt{3} , -\sqrt{3}

so, there are totally 4 real roots.