Let a function g - -0- 4- - be defined as gx-0- t- xmax t3-6t2-9t-3 - 0- x- 3 4-x - 3- x- 4 -then the number of points in the interval 0- 4 where gx is NOT differentiable- is

nbsp;f(t)=t^3 6t^2+9t 3 f'(t)=3(t 1)(t 3) Local max at x = 1, f(1) = 1 g(x)= nbsp; amp; f (x)=x^3 6x^2+9x 3 amp;x∈ [0,1] nbsp; nbsp; amp; 1 ...

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Standard XII

Mathematics

Extrema

Let a functio...

Question

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

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Solution

$f (t) = t^{3} - 6 t^{2} + 9 t - 3$

$f^{'} (t) = 3 (t - 1) (t - 3)$
Local max at $x = 1, f (1) = 1$

$g (x) = ⎧ ⎨ ⎩ \begin{matrix} f (x) = x^{3} - 6 x^{2} + 9 x - 3 & x \in [0, 1] 1 & x \in (1, 3] 4 - x & 3 < x \leq 4 \end{matrix}$

Clearly, at $x = 1$ , we have $L H D = R H D = 0$
Hence, not differentiable at $x = 3$

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Let a function g:[0,4]→R be defined as g(x)={max0≤t≤x{t3−6t2+9t−3},0≤x≤34−x,3<x≤4, then the number of points in the interval (0,4) where g(x) is NOT differentiable, is

f(t)=t3−6t2+9t−3 f′(t)=3(t−1)(t−3) Local max at x=1,f(1)=1 g(x)=⎧⎨⎩f(x)=x3−6x2+9x−3x∈[0,1] 1 x∈(1,3]4−x3<x≤4 Clearly, at x=1, we have LHD=RHD=0 Hence, not differentiable at x=3

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