The function f(x) = begin{cases} x - 1, & x

Question

(A) 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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For all real values of $x$

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(A) 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: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (x - 1) = 2 - 1 = 1$. $2$. Right-hand limit: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (2x - 3) = 2(2) - 3 = 4 - 3 = 1$. $3$. Value of the function at $x = 2$: $f(2) = 2(2) - 3 = 1$. Since $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2) = 1$,the function is continuous at $x = 2$. For $x For $x > 2$,$f(x) = 2x - 3$,which is also a polynomial function and thus continuous. Therefore,the function $f(x)$ is continuous for all real values of $x$.

The function $f(x) = \begin{cases} x - 1, & x < 2 \\ 2x - 3, & x \ge 2 \end{cases}$ is a continuous function:

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