At the point x = 1,the given function f(x) = | vedclass

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(B) We check the continuity at $x = 1$:$f(1) = 1 - 1 = 0$$LHL = \lim_{x 	o 1^-} f(x) = \lim_{h 	o 0} f(1 - h) = \lim_{h 	o 0} ((1 - h) - 1) = 0$$RH

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Continuous and not differentiable

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(B) We check the continuity at $x = 1$:$f(1) = 1 - 1 = 0$$LHL = \lim_{x 	o 1^-} f(x) = \lim_{h 	o 0} f(1 - h) = \lim_{h 	o 0} ((1 - h) - 1) = 0$$RHL = \lim_{x 	o 1^+} f(x) = \lim_{h 	o 0} f(1 + h) = \lim_{h 	o 0} ((1 + h)^3 - 1) = 0$Since $LHL = RHL = f(1)$,the function is continuous at $x = 1$.Now,we check the differentiability at $x = 1$:$Lf'(1) = \lim_{h 	o 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h 	o 0} \frac{(1 - h - 1) - 0}{-h} = \lim_{h 	o 0} \frac{-h}{-h} = 1$$Rf'(1) = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h 	o 0} \frac{((1 + h)^3 - 1) - 0}{h} = \lim_{h 	o 0} \frac{1 + 3h + 3h^2 + h^3 - 1}{h} = \lim_{h 	o 0} (3 + 3h + h^2) = 3$Since $Lf'(1) 
eq Rf'(1)$,the function is not differentiable at $x = 1$.Thus,the function is continuous and not differentiable at $x = 1$.

At the point $x = 1$,the given function $f(x) = \begin{cases} x^3 - 1; & 1 < x < \infty \\ x - 1; & -\infty < x \le 1 \end{cases}$ is

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