Let f(x) = begin{cases} x^{3/5} & text{if } x | vedclass

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(C) critical point occurs where $f'(x) = 0$ or where $f'(x)$ is undefined.For $x < 1$,$f'(x) = \frac{3}{5}x^{-2/5} = \frac{3}{5x^{2/5}}$. $f'(x)$ is u

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$3$

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(C) critical point occurs where $f'(x) = 0$ or where $f'(x)$ is undefined.For $x For $x > 1$,$f'(x) = -3(x - 2)^2$. Setting $f'(x) = 0$ gives $x = 2$. Thus,$x = 2$ is a critical point.At $x = 1$,we check for continuity and differentiability:$f(1) = 1^{3/5} = 1$.$\lim_{x 	o 1^-} f(x) = 1$ and $\lim_{x 	o 1^+} f(x) = -(1 - 2)^3 = 1$. So $f(x)$ is continuous at $x = 1$.Left-hand derivative at $x = 1$: $f'(1^-) = \frac{3}{5(1)^{2/5}} = \frac{3}{5}$.Right-hand derivative at $x = 1$: $f'(1^+) = -3(1 - 2)^2 = -3$.Since $f'(1^-) 
eq f'(1^+)$,$f(x)$ is not differentiable at $x = 1$. Thus,$x = 1$ is a critical point.The critical points are $x = 0, 1, 2$. Therefore,there are $3$ critical points.

Column $I$	Column $II$
$A$. $x\|x\|$	$I$. Strictly increasing and continuous in $(-1,1)$
$B$. $\sqrt{\|x\|}$	$II$. Continuous but not differentiable in $(-1,1)$
$C$. $x+[x]$	$III$. Differentiable in $(-1,1)$
$D$. $\|x-1\|+\|x+1\|+\|x\|$	$IV$. Differentiable in $(-1,0) \cup (0,1)$
	$V$. Strictly increasing and not differentiable in $(-1,1)$

Let $f(x) = \begin{cases} x^{3/5} & \text{if } x \le 1 \\ -(x - 2)^3 & \text{if } x > 1 \end{cases}$. Then the number of critical points on the graph of the function is:

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