Question

Let $f\left(x\right)=x^2-2x,x\in R$ and $g\left(x\right)=f\left(f\right(x)-1)+f(5-f(x\left)\right)$,show that $g\left(x\right)\ge 0,\forall x\in R$

Answer 1

$f\left(x\right)=x^2-2x$ ....(i)
and $g\left(x\right)=f\left(f\right(x)-1)+f(5-f(x\left)\right)$ ....(ii)
$\therefore g\left(x\right)=f(x^2-2x-1)+f(5-x^2+2x)$
$=\left[\right(x^2-2x-1{)}^2-2(x^2-2x-1)]+[(5-x^2+2x{)}^2-2(5-x^2+2x\left)\right]$
$\Rightarrow g\left(x\right)=2x^4-8x^3-4x^2+24x+18$

Now, we will show $g\left(x\right)\ge 0$
$g^{\prime }\left(x\right)=8x^3-24x^2-8x+24$

Let $g^{\prime }\left(x\right)=0$
$\Rightarrow x=-1,1$ and 3
$\Rightarrow g\left(x\right)\ge min\left\{g\right(-1),g(1),g(3\left)\right\}=0$
Hence, $g\left(x\right)\ge 0\forall x\in R$