Let a function g:[0,4]R be defined as g(x)=cases_0 t x(t^3-6t^2+9t-3), & 0 x34-x, & 3

Question

Let a function $g:[0,4]	o\mathbb{R}$ be defined as $g(x)=\begin{cases}\max\limits_{0\le t\le x}(t^3-6t^2+9t-3), & 0\le x\le3\4-x, & 3<x\le4\end{cases}$. Then the number of points in $(0,4)$ where $g(x)$ is not differentiable, is ......

Answer: 1

Accepted Answer

Let $h(t)=t^3-6t^2+9t-3$; $h'(t)=3(t-1)(t-3)=0$ at $t=1,3$. $h(1)=1$, $h(3)=-3$.

$$g(x)=\begin{cases}h(x), & 0\le x\le1\1, & 1\le x<3\4-x, & 3<x\le4\end{cases}$$
This is continuous. $g'(x)=3(x-1)(x-3)$ on $[0,1]$, $0$ on $(1,3)$, $-1$ on $(3,4)$. At $x=3$: LHD $=0$, RHD $=-1$ --- not equal. So $g$ is non-differentiable at exactly 1 point ($x=3$).

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