Let g(x)={[(x-2)^(2)," for "x 2]:} Is g continuous at x=2 ? Choose 1 answer: (A) Yes (B) No

Check Left-hand Limit: To determine if g(x) is continuous at x=2, we need to check if the left-hand limit, the right-hand limit, and the value of the function at x=2 are all equal.Calculate Left-hand Limit: First, let's find the left-hand limit as x approaches 2 from the left, which means we use the piece of the function defined for x 2-) g(x) = lim(x->2-) (x-2)^2.Check Right-hand Limit: Calculating the left-hand limit: lim(x->2-) (x-2)^2 = (2-2)^2 = 0.Calculate Right-hand Limit: Now, let's find the right-hand limit as x approaches 2 from the right, which means we use the piece of the function defined for x > 2: lim(x->2+) g(x) = lim(x->2+) (2-x^2).Compare Limits: Calculating the right-hand limit: lim(x->2+) (2-x^2) = 2-(2^2) = 2-4 = -2.Check Value at x=2: Since the left-hand limit (0) is not equal to the right-hand limit (-2), g(x) is not continuous at x=2.Calculate Value at x=2: Finally, let's check the value of the function at x=2. We can use either piece of the function since they should agree at x=2 if the function is continuous. Let's use the first piece: g(2) = (2-2)^2 = 0.Final Conclusion: The value of the function at x=2 (0) is equal to the left-hand limit (0) but not equal to the right-hand limit (-2). Therefore, g(x) is not continuous at x=2.

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$Let g(x)={[(x-2)^(2)," for "x <= 2],[2-x^(2)," for "x > 2]:} Is g continuous at x=2 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ $g(x)=\left\{\begin{array}{ll} (x-2)^{2} & \text { for } x \leq 2 \\ 2-x^{2} & \text { for } x>2 \end{array}\right.$ $\newline$ Is $g$ continuous at $x=2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) $\mathrm{No}$

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Q. Let

\newline

g(x)=\left\{\begin{array}{ll} (x-2)^{2} & \text { for } x \leq 2 \\ 2-x^{2} & \text { for } x>2 \end{array}\right.

\newline

g

continuous at

x=2

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Check Left-hand Limit: To determine if $g(x)$ is continuous at $x=2$ , we need to check if the left-hand limit, the right-hand limit, and the value of the function at $x=2$ are all equal.
Calculate Left-hand Limit: First, let's find the left-hand limit as $x$ approaches $2$ from the left, which means we use the piece of the function defined for $x \leq 2$ : $\lim_{x\to 2^-} g(x) = \lim_{x\to 2^-} (x-2)^2$ .
Check Right-hand Limit: Calculating the left-hand limit: $\lim_{x\to 2^-} (x-2)^2 = (2-2)^2 = 0$ .
Calculate Right-hand Limit: Now, let's find the right-hand limit as $x$ approaches $2$ from the right, which means we use the piece of the function defined for x > 2: $\lim_{x\to 2^+} g(x) = \lim_{x\to 2^+} (2-x^2)$ .
Compare Limits: Calculating the right-hand limit: $\lim_{x\to 2^+} (2-x^2) = 2-(2^2) = 2-4 = -2$ .
Check Value at $x=2$ : Since the left-hand limit ( $0$ ) is not equal to the right-hand limit ( $-2$ ), $g(x)$ is not continuous at $x=2$ .
Calculate Value at $x=2$ : Finally, let's check the value of the function at $x=2$ . We can use either piece of the function since they should agree at $x=2$ if the function is continuous. Let's use the first piece: $g(2) = (2-2)^2 = 0$ .
Final Conclusion: The value of the function at $x=2$ ( $0$ ) is equal to the left-hand limit ( $0$ ) but not equal to the right-hand limit ( $-2$ ). Therefore, $g(x)$ is not continuous at $x=2$ .