Question:
Choose the correct alternative in the following:
If $f(x)=|x-3|$ and $g(x)=$ fof $(x)$, then for $x>10, g^{\prime}(x)$ is equal to
A. 1
B. $-1$
C. 0
D. none of these
Solution:
$g(x)=f \circ f(x)=f(f(x))=|f(x)-3| \because f(x)=|x-3|$
$=|| x-3|-3|$
$\because|x-3|=\left\{\begin{array}{c}(x-3), x>3 \\ -(x-3), x<3\end{array}\right.$
Since we have given $x>10$ then $|x-3|=(x-3)$
$\therefore g(x)=|(x-3)-3|=|x-6|$
$\because|x-6|=\left\{\begin{array}{l}(x-6), x>6 \\ -(x-6), x<6\end{array}\right.$
Since we have given $x>10$ then $|x-6|=(x-6)$
$\therefore g(x)=(x-6)$
$g^{\prime}(x)=\frac{d}{d x}(x-6)=1=(A)$