Practicing Success
Let $f(x)=x^3-x^2+x+1$, $g(x)=\left\{\begin{matrix}max.\{f(t),0≤t≤x\},&0≤x≤1\\3-x,&1<x≤2\end{matrix}\right.$ Then in [0, 2] the points where g(x) is not differentiable is/are |
1 2 1 and 2 none of these |
1 |
$f(t)=t^3-t^2+t+1$ $∴f'(t)=3t^2-2t+1>0$ ∴ 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. |