Let $f(x)=x^3-x^2+x+1$ and g(x) be a function defined by

$g(x)=\left\{\begin{array}{cc} Max ~\{f(t): 0 \leq t \leq x\}, & 0 \leq x \leq 1 \\ 3-x ~~~~, & 1<x \leq 2 \end{array}\right.$

Then, g(x) is

continuous but not differentiable on [0, 2]

We have,

$f(x)=x^3-x^2+x+1$

$f'(x)=3 x^2-2 x+1>0$ for all x                     [∵ Disc < 0 and Coeff. of x² > 0]

⇒ f(x) is increasing for all $x \in R$.

∴  $g(x)=\left\{\begin{array}{cc} x^3-x^2+x+1, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2 \end{array}\right.$

Clearly, g(x) is everywhere continuous and differentiable except possibility at x = 1

Clearly, $\lim\limits_{x \rightarrow 1^{-}} g(x)=\lim\limits_{x \rightarrow 1^{+}} g(x)=g(1)$.

So, it is continuous at x = 1.

We observe that

(LHD at x = 1)

$=\left\{\frac{d}{d x}\left(x^3-x^2+x+1\right)\right\}_{x=1}$

$=\left(3 x^2-2 x+1\right)_{\text {at } x=1}=2$

and, (RHD at x = 1) = $\left(\frac{d}{d x}(3-x)\right)_{x=1}=-1$.

Clearly, g(x) is not differentiable at x = 1.