If $f(x) = |x - 1|$ and $g(x)=f(f(f(x)))$, then for x > 2, g'(x) is equal to

-1 for 2 < x < 3

We have, $f(x)=|x-1|$

∴ $g(x)=f(f(f(x)))$

$\Rightarrow g(x)=f(f(|x-1|))$

$\Rightarrow g(x)=f(|| x-1|-1|)$

$\Rightarrow g(x)=|||x-1|-1|-1|$

$\Rightarrow g(x)=\left\{\begin{array}{l} || x-2|-1|, \text { if } x \geq 1 \\ ||-x|-1|, \text { if } x<1
\end{array}\right.$

$\Rightarrow g(x)= \begin{cases}|x-3|, & \text { if } x \geq 2 \\ |2-x-1|, & \text { if } 1 \leq x<2 \\ |x-1|, & \text { if } 0 \leq x<1 \\ |-x-1|, & \text { if } x<0\end{cases}$

$\Rightarrow g(x)=\left\{\begin{array}{lll}x-3, & \text { if } \quad x \geq 3 \\ 3-x, & \text { if } 2 \leq x<3 \\ x-1, & \text { if } 1 \leq x<2 \\ 1-x, & \text { if } 0 \leq x<1 \\ x+1, & \text { if }-1 \leq x<0 \\ -x-1, & \text { if } x<-1\end{array}\right.$

$\Rightarrow g'(x)= \begin{cases}1, & \text { if } x>3 \\ -1, & \text { if } 2<x<3 \\ 1, & \text { if } 1<x<2 \\ -1, & \text { if } 0<x<1 \\ 1, & \text { if }-1<x<0 \\ -1, & \text { if } x<-1\end{cases}$

ALITER For x > 2, we have

$f(x)=|x-1|=x-1$

∴  $f(f(x))=f(x-1)=|(x-1)-1|$

$\Rightarrow f(f(x))=|x-2|=x-2$                 [∵ x > 2]

$\Rightarrow f(f(f(x)))=f(x-2)=|x-2-1|= \begin{cases}x-3, & x \geq 3 \\ 3-x, & 2 \leq x<3\end{cases}$

$\Rightarrow g(x)= \begin{cases}x-3, & x \geq 3 \\ 3-x, & 2 \leq x<3\end{cases} $

$\Rightarrow g'(x)=\left\{\begin{aligned} 1, & ~~~x>3 \\ -1, & ~~~2<x<3\end{aligned}\right.$