If f(x) = min {1, x^2, x^3},then | vedclass

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(C) To find $f(x) = \min \{1, x^2, x^3\}$,we analyze the intersection points of the functions $y=1$,$y=x^2$,and $y=x^3$.$1$. Intersection of $x^2$ and

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$f(x)$ is not differentiable but continuous $\forall \, x \in R$

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(C) To find $f(x) = \min \{1, x^2, x^3\}$,we analyze the intersection points of the functions $y=1$,$y=x^2$,and $y=x^3$.$1$. Intersection of $x^2$ and $x^3$: $x^2 = x^3 \implies x^2(x-1) = 0 \implies x=0, x=1$.$2$. Intersection of $x^2$ and $1$: $x^2 = 1 \implies x = \pm 1$.$3$. Intersection of $x^3$ and $1$: $x^3 = 1 \implies x = 1$.Analyzing the intervals:- For $x - For $-1 \le x - For $0 \le x - For $x \ge 1$,$1 \le x^2 \le x^3$,so $f(x) = 1$.Thus,$f(x) = \begin{cases} x^3, & x Checking continuity at $x=1$: $\lim_{x 	o 1^-} f(x) = 1^3 = 1$ and $f(1) = 1$. So,$f(x)$ is continuous at $x=1$.Checking differentiability at $x=1$: $f'(1^-) = \frac{d}{dx}(x^3)|_{x=1} = 3(1)^2 = 3$. $f'(1^+) = \frac{d}{dx}(1)|_{x=1} = 0$. Since $3 
eq 0$,$f(x)$ is not differentiable at $x=1$.Thus,$f(x)$ is continuous everywhere but not differentiable at $x=1$.

If $f(x) = \min \{1, x^2, x^3\}$,then

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