Question

If $f\left(x\right)=|x-2|$ and $g\left(x\right)=f\left(f\left(x\right)\right)$, then $g^{\prime }\left(x\right)$ for $x>2$ is

Answer 1

We have, $f\left(x\right)=2-x,x<2=x-2,x\ge 2$

Thus, we have,

$g\left(x\right)=f\left\{f\left(x\right)\right\}=2-f,f<2$

$=f-2,f\ge 2$

i.e., $g\left(x\right)=2-(2-x),2-x<2x<2$

$=(2-x)-2,2-x\ge 2x<2$

$=2-(x-2),x-2<2x\ge 2$

$=(x-2),-2,x-2\ge 2x\ge 2$

i.e., $g\left(x\right)=x,0<x<2$

$=-x,x\le 0$

$=4-x,2\le x<4$

$=x-4,4\le x$

i.e., $g\left(x\right)=-x,x\le 0$

$=x,0<x<2$

$=4-x,2\le x<4$

$=x-4,4\le x$

Hence, we have for $x>2$

$g^{\prime }\left(x\right)=-1,2<x<4$

$=1,4\le x$

The derivative of $g\left(x\right)$ does not exist at $x=4.$