Let f(x) = 15 – |x – 10|; x ∈ R . Then the set of all values of x ,...

f\left(x\right) = 15 - \left|x - 10\right|$</p><p>$⇒ g\left(x\right) = f\left(f\left(x\right)\right)$</p><p>$⇒ g\left(x\right) = 15 - \left|f\left(x\right) - 10\right|$</p><p>$= 15 - \left|15 - \left|x - 10\right| - 10\right|$</p><p>$= 15 - \left|5 - \left|x - 10\right|\right|$</p><p>$= 15 - \left|\left|x - 10\right| - 5\right|</p><p>We know that, \left|x\right| = \left\{\begin{cases}x, & x \ge 0 \\ - x, & x < 0\end{cases}\right)</p><p>Case1:$</p><p>When$x < 10$</p><p>$g\left(x\right) = 15 - \left|10 - x - 5\right|$</p><p>$⇒ g ( x ) = 15 - \left|5 - x\right|$</p><p>Now if$x < 5$, then$g\left(x\right) = 15 - 5 + x = x + 10$</p><p>And if$x \ge 5$, then$g ( x ) = 15 + 5 - x = 20 - x$</p><p>Case$2: x \ge 10$</p><p>$g\left(x\right) = 15 - \left|x - 10 - 5\right|$</p><p>$⇒ g ( x ) = 15 - \left|x - 15\right|$</p><p>Now if$10 \le x < 15$</p><p>$⇒ g\left(x\right) = 15 - 15 + x$</p><p>So,$g\left(x\right) = x$</p><p>Now for$x \ge 15$, then$g\left(x\right) = 15 - x + 15 = 30 - x$</p><p> <br />$g\left(x\right) = \left\{$\begin{cases}x + 10, & x < 5 \\ 20 - x, & 5 \le x < 10 \\ x, & 10 < x \le 15 \\ 30 - x, & x \ge 15\end{cases}$\right)

Now, it is clear from the graph of $g\left(x\right)$ has sharp corners at$x = 5, 10, 15$ and hence this function is not differentiable at $x = 5, 10, 15 .$