Let a function g: [0, 4] → R be defined as g(x) = \ cases max\ t ^...

We have,

g\left(x\right) = \left\{\begin{cases}\mathop{max\left\{{t}^{3} - 6{t}^{2} + 9t - 3\right\}}_{0 \le t \le x}, & 0 \le x \le 3 \\ 4 - x, & 3 < x \le 4\end{cases}\right)</p><p>Let</p><p>f\left(x\right) = {x}^{3} - 6{x}^{2} + 9x - 3$</p><p>$⇒ {f}^{'} ( x ) = 3{x}^{2} - 12x + 9$</p><p>$⇒ {f}^{'} ( x ) = 3 ( x - 1 ) ( x - 3 )$</p><p>For critical points:</p><p>${f}^{'}\left(x\right) = 0$</p><p>$⇒ x = 1, 3$</p><p>And,</p><p>${f}^{"}\left(x\right) = 6x - 12$</p><p>Then,</p><p>${f}^{"}\left(1\right) = 6 - 12 = - 6 < 0$(maxima)</p><p>${f}^{"}\left(3\right) = 6 > 0$(minima)</p><p>Hence,</p><p>$g\left(x\right) = \left[$\begin{cases}f ( x ) , & 0 \le x \le 1 \\ 1, & 1 \le x \le 3 \\ 4 - x, & 3 < x \le 4\end{cases}$\right)

Hence, $g$\left(x\right)$is continuous function.</p><p>${g}^{'}\left(x\right) = \left[$\begin{cases}3 ( x - 1 ) ( x - 3 ) , & 0 < x < 1 \\ 0, & 1 < x < 3 \\ - 1, & 3 < x < 4\end{cases}$\right)

Clearly, $g ( x )$ is non-differentiable at $x = 3$.