Question

Let $f\left(x\right)=x^3-3x+1.$ Then

• A. $f\left(f\right(x\left)\right)=0$ has $7$ real solutions
• B. $f\left(f\right(x\left)\right)=0$ has $4$ real solutions
• C. $f\left(f\right(x\left)\right)=-1$ has $7$ real solutions
• D. $f\left(f\right(x\left)\right)=-1$ has $4$ real solutions

Answer 1

Answer: (A), (D)

$f\left(f\right(x\left)\right)=-1$ has $4$ real solutions

$f\left(x\right)=x^3-3x+1$
$\Rightarrow f^{\prime }\left(x\right)=3(x-1)(x+1)$
$\Rightarrow f$ is increasing in $(-\infty ,-1]\cup [1,\infty )$ and decreasing in $[-1,1]$
$f(-2)=-1,f(-1)=3,f\left(1\right)=-1,f\left(2\right)=3$
Clearly, by intermediate value theorem,
$f\left(x\right)=0$ has three distinct roots $x_1,x_2,x_3$ i.e.,
$-2<x_1<-1<x_2<1<x_3<2$
So $f\left(x\right)=x_1$ has $1$ solution
$f\left(x\right)=x_2$ has $3$ solutions
$f\left(x\right)=x_3$ has $3$ solutions

Similarly, $f\left(f\right(x\left)\right)=-1$ has $4$ real solutions