Let g(x)={[5-x," for "x 3]:} Is g continuous at x=3 ? Choose 1 answer: (A) Yes (B) No

Check Left-Hand Limit: To determine if g(x) is continuous at x=3, we need to check if the left-hand limit, the right-hand limit, and the value of the function at x=3 are all equal.Check Right-Hand Limit: First, we find the left-hand limit as x approaches 3 from the left, which means we use the piece of the function defined for x 3-) g(x) = 5 - 3 = 2.Check Function Value at x=3: Next, we find the right-hand limit as x approaches 3 from the right, which means we use the piece of the function defined for x > 3. We substitute x=3 into the second piece of the function: lim (x -> 3+) g(x) = 2e^(3-3) = 2e^0 = 2.Conclusion: Now, we check the value of the function at x=3. Since x=3 falls in the first piece of the function (x <= 3), we use the first piece to find g(3): g(3) = 5 - 3 = 2.Since the left-hand limit, the right-hand limit, and the value of the function at x=3 are all equal to 2, the function g(x) is continuous at x=3.

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$Let g(x)={[5-x," for "x <= 3],[2e^(3-x)," for "x > 3]:} Is g continuous at x=3 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ $g(x)=\left\{\begin{array}{ll}5-x & \text { for } x \leq 3 \\ 2 e^{3-x} & \text { for } x>3\end{array}\right.$ $\newline$ Is $g$ continuous at $x=3$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) $\mathrm{No}$

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Q. Let

\newline

g(x)=\left\{\begin{array}{ll}5-x & \text { for } x \leq 3 \\ 2 e^{3-x} & \text { for } x>3\end{array}\right.

\newline

g

continuous at

x=3

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Check Left-Hand Limit: To determine if $g(x)$ is continuous at $x=3$ , we need to check if the left-hand limit, the right-hand limit, and the value of the function at $x=3$ are all equal.
Check Right-Hand Limit: First, we find the left-hand limit as $x$ approaches $3$ from the left, which means we use the piece of the function defined for $x \leq 3$ . We substitute $x=3$ into the first piece of the function: $\newline$ $\lim_{x \to 3^-} g(x) = 5 - 3 = 2.$
Check Function Value at x= $3$ : Next, we find the right-hand limit as $x$ approaches $3$ from the right, which means we use the piece of the function defined for x > 3. We substitute $x=3$ into the second piece of the function: $\newline$ $\lim_{x \to 3^+} g(x) = 2e^{3-3} = 2e^0 = 2.$
Conclusion: Now, we check the value of the function at $x=3$ . Since $x=3$ falls in the first piece of the function ( $x \leq 3$ ), we use the first piece to find $g(3)$ : $g(3) = 5 - 3 = 2.$
Conclusion: Now, we check the value of the function at $x=3$ . Since $x=3$ falls in the first piece of the function ( $x \leq 3$ ), we use the first piece to find $g(3)$ : $g(3) = 5 - 3 = 2.$ Since the left-hand limit, the right-hand limit, and the value of the function at $x=3$ are all equal to $2$ , the function $g(x)$ is continuous at $x=3$ .