MCQ
Let a function $g:[0,4] \rightarrow R$ be defined as

$g ( x )=\left\{\begin{array}{ll}\max _{0 \leq t \leq x }\left\{ t ^{3}-6 t ^{2}+9 t -3\right\} & , 0 \leq x \leq 3 \\ 4- x & , 3 < x \leq 4\end{array}\right.$ then the number of points in the interval $(0,4)$ where $g(x)$ is NOT differentiable, is $.....$

  • A
    $5$
  • B
    $3$
  • $1$
  • D
    $11$

Answer

Correct option: C.
$1$
c
$f(X)=x^{3}-6 x^{2}+9 x-3$

$f(x)=3 x^{2}-12 x+9=3(x-1)(x-3)$

$f(1)=1$ and $f(3)=-3$

$g(x)=f(x) \quad\quad 0 \leq x \leq 1$

$\quad\quad\quad\quad1 \quad\quad\quad 1 \leq x \leq 3$

$\quad\quad\quad\quad4-x \quad 3$

$g(x)$ is continuous

$g^{\prime}(x)=3(x-1)(x-3) \quad\quad 0 \leq x \leq 1$

$\quad\quad\quad\quad0\quad\quad\quad\quad \quad \quad \quad 1 \leq x \leq 3$

$\quad\quad\quad\quad-1 \quad \quad \quad \quad \quad \quad 3$

$g(x)$ is non-differentiable at $x=3$

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