The function f(x) = begin{cases} x + 2, & 1 l | vedclass

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(C) To check the continuity of the function $f(x)$ at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2

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$x \ge 2$

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(C) To check the continuity of the function $f(x)$ at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-hand limit $(LHL)$ at $x = 2$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (x + 2) = 2 + 2 = 4$.$2$. Right-hand limit $(RHL)$ at $x = 2$: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (3x - 2) = 3(2) - 2 = 6 - 2 = 4$.$3$. Value of the function at $x = 2$: $f(2) = 4$.Since $	ext{LHL} = 	ext{RHL} = f(2) = 4$,the function is continuous at $x = 2$.For $x > 2$,$f(x) = 3x - 2$,which is a polynomial function and therefore continuous for all $x > 2$.Combining these,the function is continuous for all $x \ge 2$ within its domain.Thus,the function is continuous at $x \ge 2$.

The function $f(x) = \begin{cases} x + 2, & 1 \le x \le 2 \\ 4, & x = 2 \\ 3x - 2, & x > 2 \end{cases}$ is continuous at

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