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

Check Left Limit: To determine if h(x) is continuous at x=1, we need to check if the left-hand limit as x approaches 1, the right-hand limit as x approaches 1, and the value of the function at x=1 are all equal.Calculate Left Limit: First, we calculate the left-hand limit, which is the limit of h(x) as x approaches 1 from the left. Since x 1-) 2^(x-1) = 2^(1-1) = 2^0 = 1.Calculate Right Limit: Next, we calculate the right-hand limit, which is the limit of h(x) as x approaches 1 from the right. Since x >= 1, we use the second part of the function definition: h(x) = 2^(1-x). So, we need to find the limit of 2^(1-x) as x approaches 1 from the right. lim (x -> 1+) 2^(1-x) = 2^(1-1) = 2^0 = 1.Find Value at x=1: Now, we need to find the value of the function at x=1. Since x=1 is the point where the definition of the function changes, we need to check both parts of the function definition. For x = 1, h(x) = 2^(1-x), so h(1) = 2^(1-1) = 2^0 = 1. Both parts of the function give the same value at x=1, which is 1.Check Continuity: Since the left-hand limit, the right-hand limit, and the value of the function at x=1 are all equal to 1, the function h(x) is continuous at x=1.

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

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

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Q. Let

\newline

h(x)=\left\{\begin{array}{ll} 2^{x-1} & \text { for } x<1 \\ 2^{1-x} & \text { for } x \geq 1 \end{array}\right.

\newline

h

continuous at

x=1

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check Left Limit: To determine if $h(x)$ is continuous at $x=1$ , we need to check if the left-hand limit as $x$ approaches $1$ , the right-hand limit as $x$ approaches $1$ , and the value of the function at $x=1$ are all equal.
Calculate Left Limit: First, we calculate the left-hand limit, which is the limit of $h(x)$ as $x$ approaches $1$ from the left. Since x < 1, we use the first part of the function definition: $h(x) = 2^{(x-1)}$ . $\newline$ So, we need to find the limit of $2^{(x-1)}$ as $x$ approaches $1$ from the left. $\newline$ $\lim_{x \to 1^-} 2^{(x-1)} = 2^{(1-1)} = 2^0 = 1$ .
Calculate Right Limit: Next, we calculate the right-hand limit, which is the limit of $h(x)$ as $x$ approaches $1$ from the right. Since $x \geq 1$ , we use the second part of the function definition: $h(x) = 2^{(1-x)}$ . $\newline$ So, we need to find the limit of $2^{(1-x)}$ as $x$ approaches $1$ from the right. $\newline$ $\lim_{x \to 1^+} 2^{(1-x)} = 2^{(1-1)} = 2^0 = 1$ .
Find Value at $x=1$ : Now, we need to find the value of the function at $x=1$ . Since $x=1$ is the point where the definition of the function changes, we need to check both parts of the function definition. $\newline$ For x < 1, $h(x) = 2^{(x-1)}$ , so $h(1) = 2^{(1-1)} = 2^0 = 1$ . $\newline$ For $x \geq 1$ , $h(x) = 2^{(1-x)}$ , so $h(1) = 2^{(1-1)} = 2^0 = 1$ . $\newline$ Both parts of the function give the same value at $x=1$ , which is $x=1$ $0$ .
Check Continuity: Since the left-hand limit, the right-hand limit, and the value of the function at $x=1$ are all equal to $1$ , the function $h(x)$ is continuous at $x=1$ .