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(D) To find the points where $f(x) = \max(x, x^3)$ is not differentiable,we analyze the behavior of $x$ and $x^3$ in different intervals:$1$. If $x <

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$\{ - 1, 0, 1\}$

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(D) To find the points where $f(x) = \max(x, x^3)$ is not differentiable,we analyze the behavior of $x$ and $x^3$ in different intervals:$1$. If $x $2$. If $-1  x$,so $f(x) = x^3$.$3$. If $0  x^3$,so $f(x) = x$.$4$. If $x > 1$,then $x^3 > x$,so $f(x) = x^3$.At the transition points $x = -1, 0, 1$,we check the left-hand and right-hand derivatives:At $x = -1$: $f'( -1^-) = 1$ and $f'( -1^+) = 3(-1)^2 = 3$. Since $1 
eq 3$,it is not differentiable.At $x = 0$: $f'( 0^-) = 3(0)^2 = 0$ and $f'( 0^+) = 1$. Since $0 
eq 1$,it is not differentiable.At $x = 1$: $f'( 1^-) = 1$ and $f'( 1^+) = 3(1)^2 = 3$. Since $1 
eq 3$,it is not differentiable.Thus,the set of points where $f(x)$ is not differentiable is $\{ -1, 0, 1\}$.

Let $f:R \to R$ be a function defined by $f(x) = \max \,(x, x^3).$ The set of all points where $f(x)$ is not differentiable is

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