Question

Let $f\left(x\right)=(x+1{)}^2-1,(x\ge -1)$. Then the set $S=\{x:f(x)=f^{-1}(x\left)\right\}$ is

• A. $\{0,-1,\frac{-3+i\sqrt{3}}{2},\frac{-3-i\sqrt{3}}{2}\}$
• B. $\{0,1,-1\}$
• C. $\{0,-1\}$
• D. empty.

Answer 1

The correct option is A $\{0,-1,\frac{-3+i\sqrt{3}}{2},\frac{-3-i\sqrt{3}}{2}\}$

$f\left(x\right)=f^{-1}\left(x\right)$

$f\left(f\right(x\left)\right)=f\left(f^{-1}\right(x\left)\right)$

$f\left(f\right(x\left)\right)=x$

$f\left(x\right)=(x+1{)}^2-1$

$\therefore f\left(f\right(x\left)\right)=\left(\right((x+1{)}^2-1)+1{)}^2-1$

$\therefore f\left(f\right(x\left)\right)=(x+1{)}^4-1$

$f\left(f\right(x\left)\right)=x$

$⟹(x+1{)}^4-1=x$

$⟹(x+1{)}^4-(x+1)=0$

$⟹(x+1)\left(\right(x+1{)}^3-1)=0$

$⟹(x+1)(x^3+3x^2+3x+1-1)=0$

$⟹(x+1)\left(x\right)(x^2+3x+3)=0$

$\therefore x=0,x=-1$

$x^2+3x+3=0$

$\therefore x=\frac{-3\pm \sqrt{3^2-4\left(1\right)\left(3\right)}}{2}$

$\therefore x=\frac{-3\pm \sqrt{-3}}{2}$

$\therefore x=\frac{-3\pm \sqrt{3}i}{2}$