Question

If $f\left(x\right)=\left|x-1\right|$ and $g\left(x\right)=f\left(f\left(f\left(x\right)\right)\right)$ the value for $x>2,g\text{'}\left(x\right)=$

• A. $1$ for all $x>2$
• B. $1$ for $2<x<3$
• C. $-1$ for $2<x<3$
• D. Not defined

Answer 1

The correct option is C

$-1$ for $2<x<3$

Explanation for the correct option:

Finding the value of $g\text{'}\left(x\right)$:

Given that,

$f\left(x\right)=\left|x-1\right|$

For $x>2$

$f\left(x\right)=x-1$

Now,

$\begin{array}{rcl}f\left(f\left(x\right)\right)& =& f\left(x-1\right)\\ & =& \left|x-1-1\right|\\ & =& \left|x-2\right|\end{array}$

$\begin{array}{rcl}f\left(f\left(f\left(x\right)\right)\right)& =& f\left(x-2\right)\\ & =& \left|x-2-1\right|\\ & =& \left|x-3\right|\end{array}$

$\begin{array}{rcl}g\left(x\right)& =& f\left(f\left(f\left(x\right)\right)\right)\\ & =& \left|x-3\right|\end{array}$

For $x>3$,

$g\left(x\right)=x-3$

and for $x<3$

$\begin{array}{rcl}g\left(x\right)& =& -\left(x-3\right)\\ & =& 3-x\end{array}$

Differentiate the above function with respect to $x$.

$g\text{'}\left(x\right)=1\mathrm{and}g\mathit{\text{'}}\left(x\right)\mathit{=}-1$

Here, $g\text{'}\left(x\right)=-1$ satisifes the conditon $2<x<3$.

Hence, the correct option is C.