If $f(x)=\left\{\begin{array}{ll} \quad \frac{x^2}{2}~~~~, & 0 \leq x<1 \\ 2 x^2-3 x+\frac{3}{2}, & 1 \leq x \leq 2\end{array}\right.$, then

f and f' are continuous in [0, 2] whereas f'' is continuous in $[0, 1) \cup(1,2]$

Clearly, f is continuous and differentiable at x = 1.

Thus, f(x) is continuous and diffrentiable on [0, 2].

∴  $f'(x)= \begin{cases} \quad x, & 0 \leq x<1 \\ 4 x-3, & 1 \leq x \leq 2\end{cases}$

We observe that f'(x) is continuous on [0, 2] but it is not differentiable at x = 1.

∴  $f''(x)= \begin{cases}1, & 0 \leq x<1 \\ 4, & 1<x \leq 2\end{cases}$

Clearly, f''(x) is not continuous at x = 1.

Hence, f and f' are continuous on [0, 2] whereas f'' is continuous on $[0,1) \cup(1,2]$.