Let f(x)=x^3-x^2+x+1, g(x)=≤ft […] . Then in [0, 2] the points where…

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

$∴f'(t)=3t^2-2t+1>0$

∴ f(t) is an increasing function.

Since 0 ≤ t ≤ x

∴ max. $f(t) = f(x) = x^3-x^2+x+1$

Thus $g(x)=x^3-x^2+x+1,\,0≤x≤1$ and $g(x) = 3 − x, 1 < x ≤ 2$

The only doubtful point for differentiability of g(x) in [0, 2] is x = 1.

Clearly, $\underset{x→1^+0}{\lim}g(x)=\underset{x→1^+}{\lim}(3-x)=2$ and g(1) = 2

∴ g(x) is continuous at x = 1

Also $g'(x)=3x^2-2x+1,\,0≤x<1$ and $g'(x)=-1,\,1,x≤2$

∴ $g'(1-0)=3.1^2-2.1+1=2$ and $g'(1 + 0) = −1$

Hence g(x) is not differentiable at x = 1.