Let $f(x)=x^{3}-x^{2}+x+1$ and $g(x)= \begin{cases}\max \{f(t) \mid 0 \leq t \leq x\}, & 0 \leq x \l

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Let $f(x)=x^{3}-x^{2}+x+1$ and $g(x)= \begin{cases}\max \{f(t) \mid 0 \leq t \leq x\}, & 0 \leq x \leq 1 \ 3-x, & 1<x \leq 2\end{cases}$ \The function $g(x)$ in the interval $(0,2)$ is

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$f(x)=x^{3}-x^{2}+x+1$\ $\Rightarrow \quad f^{\prime}(x)=3 x^{2}-2 x+1$\ $f^{\prime}(x)=3 x^{2}-2 x+1=3\left(x^{2}-\frac{2}{3} x+\frac{1}{3}ight)$\ $=3\left[\left(x-\frac{1}{3}ight)^{2}-\frac{1}{9}+\frac{1}{3}ight]=3\left[\left(x-\frac{1}{3}ight)^{2}+\frac{2}{9}ight]>0$\ for all $x \in R$\ Hence $f(x)$ is an increasing function of $x$ for all real values of $x$.\ Now max $[f(t): 0 \leq t \leq x]$ the greatest value of $f(t)$ in $0 \leq t \leq x$ which is obtained at $t=x$.\ Since $f(t)$ is increasing for all $t$\ $	herefore \max [f(t): 0 \leq t \leq x]=x^{3}-x^{2}+x+1$\ Hence the function $g$ is defined as follows:\ $g(x)=x^{3}-x^{2}+x+1 \quad$ when $0 \leq x \leq 1$ $$ =3-x \quad 	ext { when } 1<x \leq 2 $$ Here R.H.L at $x=1$ is $\lim _{x ightarrow 1}(3-x)=2$\ and L.H.L at $x=1$ is $\lim _{x ightarrow 1}\left(x^{3}-x^{2}+x+1ight)=2$\ $	herefore \quad g$ is continuous at $x=1$\ Also $g^{\prime}(x)=\left\{\begin{array}{cc}3 x^{2}-2 x+1, & 0 \leq x \leq 1 \ -1, & 1<x \leq 2\end{array}ight.$\ $\operatorname{Lg}^{\prime}(1)=3(1)^{2}-2(1)+1=2$\ $R g^{\prime}(1)=-1$\ $	herefore \quad L g^{\prime}(1) 
eq R g^{\prime}(1)$\ $	herefore \quad g^{\prime}$ is not differentiable at $x=1$.\ Hence $g(x)$ is continuous on $(0,2)$. It is also derivable on $(0,2)$ except at $x=1$, where it is not differentiable.\