Let a function g:[0,4] rightarrow R be define | vedclass

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(C) Let $f(t) = t^3 - 6t^2 + 9t - 3$. Then $f'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$. The critical points are $t=1$ and $t=3$. $f(0) = -3$,$f(1) = 1 - 6

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$1$

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(C) Let $f(t) = t^3 - 6t^2 + 9t - 3$. Then $f'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$. The critical points are $t=1$ and $t=3$. $f(0) = -3$,$f(1) = 1 - 6 + 9 - 3 = 1$,and $f(3) = 27 - 54 + 27 - 3 = -3$. For $0 \leq x \leq 1$,$f(t)$ is increasing,so $\max_{0 \leq t \leq x} f(t) = f(x)$. For $1 Thus,$g(x) = \begin{cases} x^3 - 6x^2 + 9x - 3 & , 0 \leq x \leq 1 \ 1 & , 1 Now check differentiability: At $x=1$: $g'(1^-) = f'(1) = 0$ and $g'(1^+) = 0$. So $g(x)$ is differentiable at $x=1$. At $x=3$: $g(3^-) = 1$ and $g(3^+) = 4-3 = 1$. $g(x)$ is continuous at $x=3$. $g'(3^-) = 0$ and $g'(3^+) = -1$. Since $g'(3^-) 
eq g'(3^+)$,$g(x)$ is not differentiable at $x=3$. Thus,there is only $1$ point in $(0,4)$ where $g(x)$ is not differentiable.

Let a function $g:[0,4] \rightarrow R$ be defined as
$g(x) = \begin{cases} \max_{0 \leq t \leq x} \{t^3 - 6t^2 + 9t - 3\} & , 0 \leq x \leq 3 \\ 4 - x & , 3 < x \leq 4 \end{cases}$
Then the number of points in the interval $(0,4)$ where $g(x)$ is $NOT$ differentiable is $.....$

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Let a function $g:[0,4] \rightarrow R$ be defined as$g(x) = \begin{cases} \max_{0 \leq t \leq x} \{t^3 - 6t^2 + 9t - 3\} & , 0 \leq x \leq 3 \\ 4 - x & , 3 < x \leq 4 \end{cases}$Then the number of points in the interval $(0,4)$ where $g(x)$ is $NOT$ differentiable is $.....$